Introduction
The general solution of a linear differential equation with constant coefficients is the sum of the complementary function (CF) and the particular integral (PI):
\[ y(x) = y_c(x) + y_p(x) \]
General Form & Operator Notation
We write the differential equation as:
\[ f(D)y = X(x), \quad D = \frac{d}{dx} \]
Complementary Function (CF)
Solve the homogeneous equation:
\[ f(D)y = 0 \]
The auxiliary equation \( f(m) = 0 \) gives roots that decide the form of CF.
Particular Integral (PI)
A particular solution is found as:
\[ y_p = \frac{1}{f(D)}X(x) \]
Steps to Solve
- Form the auxiliary equation and find CF.
- Apply rules for PI depending on \(X(x)\).
- Write general solution: \( y = y_c + y_p \).
Complementary Function (CF) — Roots of Auxiliary Equation
The Complementary Function (CF) depends on the nature of the roots of the auxiliary (characteristic) equation:
| Case | Roots of Auxiliary Equation | Form of Complementary Function \(y_c\) | Example |
|---|---|---|---|
| 1 | Two distinct real roots \(m_1, m_2\) | \(y_c = C_1 e^{m_1 x} + C_2 e^{m_2 x}\) | \(m^2-5m+6=0 \Rightarrow m=2,3 \Rightarrow y_c=C_1 e^{2x}+C_2 e^{3x}\) |
| 2 | \(n\) distinct real roots \(m_1,\dots,m_n\) | \(y_c = \displaystyle\sum_{k=1}^n C_k e^{m_k x}\) | \(m^3-6m^2+11m-6=0 \Rightarrow m=1,2,3 \Rightarrow y_c=\sum_{k=1}^3 C_k e^{m_k x}\) |
| 3 | One pair of complex conjugate roots \(\alpha \pm i\beta\) | \(y_c = e^{\alpha x}\big(C_1\cos\beta x + C_2\sin\beta x\big)\) | \(m^2+4m+13=0 \Rightarrow m=-2\pm3i \Rightarrow y_c=e^{-2x}(C_1\cos3x + C_2\sin3x)\) |
| 4 | Mixture of distinct real and (one or more) complex roots | Combine the corresponding terms: real roots give \(C e^{mx}\); complex pairs give \(e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)\) | \(m=(1,\, -1,\, 2\pm i)\Rightarrow y_c=C_1e^{x}+C_2e^{-x}+e^{2x}(C_3\cos x + C_4\sin x)\) |
| 5 | Repeated real root of multiplicity \(r\) (e.g. \(m\) repeated) | \(y_c = (C_1 + C_2 x + C_3 x^2 + \dots + C_r x^{\,r-1}) e^{m x}\) | \(m^2-4m+4=0 \Rightarrow m=2,2 \Rightarrow y_c=(C_1+C_2x)e^{2x}\) |
| 6 | Repeated complex roots \((\alpha \pm i\beta)\) of multiplicity \(r\) | \(y_c = e^{\alpha x}\Big[(C_1 + C_2 x + \dots + C_r x^{\,r-1})\cos(\beta x) + (D_1 + D_2 x + \dots + D_r x^{\,r-1})\sin(\beta x)\Big]\) | \((m+2)^2+9=0 \Rightarrow m=-2\pm3i\) (repeated) \(y_c = e^{-2x}\big((C_1+C_2x)\cos3x+(D_1+D_2x)\sin3x\big)\) |
Worked Example
Solve \( (D^2 - 3D + 2)y = e^{2x} \):
Auxiliary equation: \( m^2 - 3m + 2 = 0 \implies m=1,2 \)
CF: \( y_c = C_1 e^x + C_2 e^{2x} \)
For PI: try \( y_p = \frac{e^{2x}}{f(2)} \). But \( f(2)=0 \), so multiply by \(x\).
\( y_p = x e^{2x} \)
General solution: \( y = C_1 e^x + C_2 e^{2x} + x e^{2x} \)
Tips & Pitfalls
- Check multiplicity of roots to adjust CF form.
- If denominator vanishes in PI, multiply trial by \(x\).
- Always verify by substitution.
Conditions for Finding Particular Integral (PI)
The method of evaluating the PI depends on the form of the non-homogeneous term \(X(x)\). Below are the standard cases:which are solved by Inverse Operator Method
Case 1: Exponential RHS (\(X(x) = e^{ax}\))
If \(f(a) \neq 0\):
\[ y_p = \frac{e^{ax}}{f(a)} \]If \(f(a) = 0\), multiply by \(x^k\) until denominator is non-zero:
\[ y_p = \frac{x^k e^{ax}}{f^{(k)}(a)} \]Case 2: Trigonometric RHS (\(X(x) = \sin bx\) or \(\cos bx\))
Replace \(D^2\) by \(-b^2\):
\[ y_p = \frac{\sin bx}{f(-b^2)}, \quad y_p = \frac{\cos bx}{f(-b^2)} \]If \(f(-b^2) = 0\), multiply trial solution by \(x\).
Case 3: Polynomial RHS (\(X(x) = P_n(x)\))
Expand the operator inverse in powers of \(D\):
\[ \frac{1}{f(D)} = A_0 + A_1 D + A_2 D^2 + \cdots \]Apply term by term on the polynomial.
Case 4: Exponential × Trigonometric (\(X(x) = e^{ax}\sin bx\) or \(e^{ax}\cos bx\))
Substitute \(D = a \pm ib\):
\[ y_p = \frac{e^{ax}(M \cos bx + N \sin bx)}{f(a \pm ib)} \]If denominator = 0, multiply by \(x\).
Case 5: Other Functions
For special cases (e.g., products, step, delta functions), use methods like:
- Method of undetermined coefficients
- Variation of parameters
Summary: The choice of PI depends entirely on the form of the RHS \(X(x)\). If the straightforward substitution leads to division by zero, multiply by suitable powers of \(x\) until a valid PI is obtained.
Finding Complementary Functions of some Differential Equations
Solve these Differential Equations
\(\left( {{D}^{2}}-2D+1 \right)y={{e}^{-x}}\)
\({{\left( D-2 \right)}^{2}}={{e}^{2x}}\)
\( \left( {{D}^{2}}+3D-10 \right)y=6{{e}^{4x}}\)
\( \left( {{D}^{2}}+1 \right)y=\cos x \)
\( \left( {{D}^{3}}+2 \right)y=\sin \left( 4x+5 \right) \)
\( \left( {{D}^{2}}-5D+6 \right)y=3\sin x \)
\( \left( {{D}^{2}}+1 \right)y=\sin x-\cos x \)
\( \left( {{D}^{2}}-1 \right)y=x\sin 3x \)
\(\left( {{D}^{4}}+2{{D}^{2}}+1 \right)y={{x}^{2}}\cos x\)
\( \left( {{D}^{2}}+1 \right)y=x{{e}^{2x}} \)
\( \left( {{D}^{2}}+5D+6 \right)y ={{e}^{x}} \)
\(\left( {{D}^{2}}+1 \right)y =\cos \left( 2x-1 \right)\)
\( \left( {{D}^{3}}+4D \right)y=\sin 2x\)
\( \left( {{D}^{2}}+D \right)y={{x}^{2}}+2x+4 \)
\( \left( {{D}^{2}}-2D+4 \right)y={{e}^{x}}\cos x \)
\( \left( {{D}^{2}}-4 \right)y=x\sinh x \)
\( \left( {{D}^{2}}-1 \right)y=x\sin 3x+\cos x\)
\( \left( {{D}^{2}}-2D+1 \right)y=x{{e}^{x}}\sin x\)
\(\left( {{D}^{2}}+{{a}^{2}} \right)y=\sec ax \)
\(\left( {{D}^{3}}-2D+1 \right)y={{e}^{x}}\log x \)
\(\left( {{D}^{2}}+1 \right)y=\cosh x \)
\(\left( {{D}^{2}}+1 \right)y=\cos ecx\)
\(\left( {{D}^{2}}+2D+1 \right)y={{e}^{-x}}\log x \)
\( \left( {{D}^{2}}+6D+9 \right)y=\frac{{{e}^{-3x}}}{{{x}^{3}}} \)
\( \left( {{D}^{2}}+3D+2 \right)y=\frac{1}{1+{{e}^{-x}}}\)
\( \left( {{D}^{2}}+1 \right)y=\frac{1}{1+\sin x} \)
\( \left( D^{2} - 2D + 1 \right) y \;=\; x e^{x} \log x \)
\( \left( {{D}^{2}}+1 \right)y=\sin x-\cos 2x \)
\( \left( {{D}^{2}}+4D+4 \right)y=2\sinh 2x \)
\( \left( {{D}^{2}}+1 \right)y=cosecx \)
\( \left( {{D}^{2}}+1 \right)y=secx \)
\( \left( {{D}^{2}}+4 \right)y=2\tan2x \)
\( \left( {{D}^{2}}+6D+9 \right)y=\frac{{{e}^{-3x}}}{{{x}^{3}}} \)
Solutons of Differential Equations
\(\left( {{D}^{2}}-2D+1 \right)y={{e}^{-x}}\)
\[\begin{align} & \left( {{D}^{2}}-2D+1 \right)y={{e}^{-x}} \\ & {{m}^{2}}-2m+1=0 \\ & \Rightarrow {{\left( m-1 \right)}^{2}}=0 \\ & \Rightarrow m=1,1 \\ & \Rightarrow {{y}_{CF}}=\left( {{c}_{1}}+{{c}_{2}}x \right){{e}^{x}} \\ & {{y}_{PI}}=\frac{{{e}^{-x}}}{{{D}^{2}}-2D+1} \\ & =\frac{{{e}^{-x}}}{{{\left( D-1 \right)}^{2}}} \\ & =\frac{{{e}^{-x}}}{{{\left( -1-1 \right)}^{2}}} \\ & =\frac{{{e}^{-x}}}{4} \\ & \Rightarrow y={{y}_{CF}}+{{y}_{PI}} \\ & \Rightarrow y=\left( {{c}_{1}}+{{c}_{2}}x \right){{e}^{x}}+\frac{{{e}^{-x}}}{4} \\ \end{align}\]\({{\left( D-2 \right)}^{2}}y={{e}^{2x}}\)
\[\begin{align} & {{\left( D-2 \right)}^{2}}y={{e}^{2x}} \\ & {{\left( m-2 \right)}^{2}}=0 \\ & \Rightarrow m=2,2 \\ & {{y}_{Cf}}=\left( {{c}_{1}}+{{c}_{2}}x \right){{e}^{2x}} \\ & {{y}_{PI}}=\frac{{{e}^{2x}}}{{{\left( D-2 \right)}^{2}}} \\ & ={{x}^{2}}\frac{{{e}^{2x}}}{{{f}^{''}}\left( a \right)} \\ & ={{x}^{2}}\frac{{{e}^{2x}}}{2} \\ & y={{y}_{Cf}}+{{y}_{PI}} \\ & =\left( {{c}_{1}}+{{c}_{2}}x \right){{e}^{2x}}+{{x}^{2}}\frac{{{e}^{2x}}}{2} \\ \end{align}\]\( \left( {{D}^{2}}+3D-10 \right)y=6{{e}^{4x}}\)
\[\begin{align} & \left( {{D}^{2}}+3D-10 \right)y=6{{e}^{4x}} \\ & \Rightarrow {{m}^{2}}+3m-10=0 \\ & \Rightarrow m=\frac{-3\pm \sqrt{9+40}}{2} \\ & \Rightarrow m=5,-2 \\ & \Rightarrow {{y}_{Cf}}={{c}_{1}}{{e}^{5x}}+{{c}_{2}}{{e}^{-2x}} \\ & {{y}_{PI}}=\frac{6{{e}^{4x}}}{{{D}^{2}}+3D-10} \\ & =\frac{6{{e}^{4x}}}{{{4}^{2}}+3\times 4-10} \\ & =\frac{6{{e}^{4x}}}{18}=\frac{{{e}^{4x}}}{3} \\ & \Rightarrow y={{c}_{1}}{{e}^{5x}}+{{c}_{2}}{{e}^{-2x}}+\frac{{{e}^{4x}}}{3} \\ \end{align}\]Method-1
Solve \( \left( {{D}^{2}}+1 \right)y=\cos x \)
\begin{align} \left( {{D}^{2}}+1 \right)y &=\cos x \\ & \Rightarrow {{m}^{2}}+1=0 \\ & \Rightarrow m=i,-i \\ \Rightarrow {{y}_{CF}}&=A\cos x+B\sin x \\ & {{y}_{PI}}=\frac{\cos x}{{{D}^{2}}+1} \\ & =x\frac{\cos x}{2D} \\ & =\frac{x}{2}\sin x \end{align}Method-2
\[\begin{align} {{y}_{PI}} &=RP\frac{{{e}^{ix}}}{{{D}^{2}}+1} \\ & =RP\,x\frac{{{e}^{ix}}}{2D} \\ & =RP\,x\frac{{{e}^{ix}}}{2\left( i \right)} \\ & =RP\,\left( \frac{x\cos x+ix\sin x}{2i} \right) \\ & =\frac{x\sin x}{2} \\ \Rightarrow y& =A\cos x+B\sin x+\frac{x}{2}\sin x \end{align}\]Methd-1 to find Particular Integral
\[ \text{Solve }(D^3+2)y=\sin(4x+5) \] \begin{align*} y &= \frac{1}{D^3+2}\,\sin(4x+5) \\[6pt] &= \Im\!\left\{\frac{1}{D^3+2}e^{i(4x+5)}\right\} \qquad\text{(since }\sin\theta=\Im\{e^{i\theta}\}\text{)}\\[6pt] &= \Im\!\left\{\frac{1}{(i4)^3+2}\;e^{i(4x+5)}\right\} \qquad\text{(replace }D\mapsto i\cdot 4\text{ for }e^{i4x}\text{)}\\[6pt] &= \Im\!\left\{\frac{1}{-64i+2}\;e^{i(4x+5)}\right\} = \Im\!\left\{\frac{1}{2-64i}\;e^{i(4x+5)}\right\} \\[6pt] &= \Im\!\left\{\frac{2+64i}{2^2+64^2}\;e^{i(4x+5)}\right\} = \Im\!\left\{\left(\frac{1}{2050}+\frac{32}{2050}i\right)e^{i(4x+5)}\right\}\\[6pt] &= \frac{1}{2050}\sin(4x+5) \;+\;\frac{32}{2050}\cos(4x+5) \\[6pt] &= \frac{1}{2050}\big(\sin(4x+5)+32\cos(4x+5)\big). \end{align*} The general solution is the sum of this particular integral and the general solution of the homogeneous equation \((D^3+2)y=0\). The characteristic equation is \(r^3+2=0\), so the roots are the cube roots of \(-2\). Let \(a=2^{1/3}\). Then the roots are \[ r_1=-a,\qquad r_{2,3}= \frac{a}{2}\pm i\frac{\sqrt3\,a}{2}. \] Hence the homogeneous solution can be written as \[ y_h=C_1e^{-a x}+e^{\tfrac{a}{2}x}\Big(C_2\cos\big(\tfrac{\sqrt3\,a}{2}x\big)+C_3\sin\big(\tfrac{\sqrt3\,a}{2}x\big)\Big). \] Finally, \[ \boxed{\,y(x)=y_h+y_p = C_1e^{-a x}+e^{\tfrac{a}{2}x}\Big(C_2\cos\big(\tfrac{\sqrt3\,a}{2}x\big)+C_3\sin\big(\tfrac{\sqrt3\,a}{2}x\big)\Big) +\frac{1}{2050}\big(\sin(4x+5)+32\cos(4x+5)\big),} \] where \(a=2^{1/3}\) and \(C_1,C_2,C_3\) are arbitrary constants.Method-2 to find Particular Integral
\[\begin{align} \Rightarrow {{y}_{PI}}&=\frac{\sin \left( 4x+5 \right)}{\left( {{D}^{3}}+2 \right)} \\ & =\frac{\sin \left( 4x+5 \right)}{\left( D{{D}^{2}}+2 \right)} \\ & =\frac{\sin \left( 4x+5 \right)}{\left( D\left( -{{4}^{2}} \right)+2 \right)} \\ & =\frac{\sin \left( 4x+5 \right)}{\left( -16D+2 \right)} \\ & =\frac{\left( -16D-2 \right)\sin \left( 4x+5 \right)}{256{{D}^{2}}-4} \\ & =\frac{\left( -16D-2 \right)\sin \left( 4x+5 \right)}{256\left( -{{4}^{2}} \right)-4} \\ & =\frac{\left( -16D-2 \right)\sin \left( 4x+5 \right)}{-4100} \\ & =\frac{\left( -16D\sin \left( 4x+5 \right)-2\sin \left( 4x+5 \right) \right)}{-4100} \\ & =\frac{\left( -64\cos \left( 4x+5 \right)-2\sin \left( 4x+5 \right) \right)}{-4100} \\ & =\frac{32}{2050}\cos \left( 4x+5 \right)+\frac{1}{2050}\sin \left( 4x+5 \right) \\ \end{align}\]Solve \(\left( {{D}^{2}}-5D+6 \right)\)y=3sinx
Method-1
\[\begin{align} & {{y}_{PI}}=\frac{3\sin x}{\left( {{D}^{2}}-5D+6 \right)} \\ & =\frac{3\sin x}{\left( -{{1}^{2}}-5D+6 \right)} \\ & =\frac{3\sin x}{-5\left( D-1 \right)} \\ & =\frac{3\left( D+1 \right)\sin x}{-5\left( {{D}^{2}}-1 \right)} \\ & =\frac{3\left( D+1 \right)\sin x}{-5\left( -{{1}^{2}}-1 \right)} \\ & =\frac{3\left( \cos x+\sin x \right)}{10} \\ \end{align}\]Method-2
\[\begin{align} & {{y}_{PI}}=3I.P\left( \frac{{{e}^{ix}}}{\left( {{D}^{2}}-5D+6 \right)} \right) \\ & =3I.P\left( \frac{{{e}^{ix}}}{\left( {{i}^{2}}-5i+6 \right)} \right) \\ & =3I.P\left( \frac{{{e}^{ix}}}{\left( -5i+5 \right)} \right) \\ & =3I.P\left( \frac{{{e}^{ix}}}{5\left( 1-i \right)} \right) \\ & =\frac{3}{5}I.P\left( \frac{{{e}^{ix}}\left( 1+i \right)}{\left( 1-i \right)\left( 1+i \right)} \right) \\ & =\frac{3}{5}I.P\left( \frac{{{e}^{ix}}\left( 1+i \right)}{\left( {{1}^{2}}-{{i}^{2}} \right)} \right) \\ & =\frac{3}{5}I.P\left( \frac{{{e}^{ix}}\left( 1+i \right)}{2} \right) \\ & =\frac{3}{10}I.P\left( \left( \cos x+i\sin x \right)\left( 1+i \right) \right) \\ & =\frac{3}{10}I.P\left( \cos x+i\cos x+i\sin x-\sin x \right) \\ & =\frac{3}{10}\left( \cos x+\sin x \right) \\ \end{align}\]\( \left( {{D}^{2}}+1 \right)y=\sin x-\cos x\)
\[\begin{align} & \left( {{D}^{2}}+1 \right)y=\sin x-\cos x \\ & {{y}_{p}}=\frac{\sin x-\cos x}{{{D}^{2}}+1} \\ & =\frac{\sin x}{{{D}^{2}}+1}-\frac{\cos x}{{{D}^{2}}+1} \\ & =x\frac{\sin x}{2D}-x\frac{\cos x}{2D} \\ & =-\frac{x}{2}\cos x-\frac{x}{2}\sin x \\ & =-\frac{x}{2}\left( \cos x+\sin x \right) \\ \end{align}\]\(\left( {{D}^{2}}-1 \right)y=x\sin3x\)
\[\begin{align} & {{y}_{p}}=\frac{x\sin 3x}{\left( {{D}^{2}}-1 \right)} \\ & =IP\left( \frac{x{{e}^{i3x}}}{\left( {{D}^{2}}-1 \right)} \right) \\ & =IP\left( {{e}^{i3x}}\frac{x}{\left( {{\left( D+3i \right)}^{2}}-1 \right)} \right) \\ & =IP\left( {{e}^{i3x}}\frac{x}{\left( {{D}^{2}}-9+6Di-1 \right)} \right) \\ & =IP\left( {{e}^{i3x}}\frac{x}{\left( {{D}^{2}}+6Di-10 \right)} \right) \\ & =IP\left( -\frac{{{e}^{i3x}}}{10}\frac{x}{\left( 1-\frac{{{D}^{2}}+6Di}{10} \right)} \right) \\ & =IP\left( -\frac{{{e}^{i3x}}}{10}\left( 1+\frac{{{D}^{2}}+6Di}{10}+{{\left( \frac{{{D}^{2}}+6Di}{10} \right)}^{2}}+... \right)x \right) \\ & =IP\left( -\frac{{{e}^{i3x}}}{10}\left( x+\frac{3}{5}i \right) \right) \\ & =IP\left( -\frac{{{e}^{i3x}}}{10}\left( x+\frac{3}{5}i \right) \right) \\ & =-IP\left( \frac{\cos 3x+i\sin 3x}{10}\left( x+\frac{3}{5}i \right) \right) \\ & =-\frac{1}{10}\left( \frac{3\cos 3x}{5}+x\sin 3x \right) \\ \end{align}\]Method-2
\[\begin{align} & {{y}_{P}}=\frac{x\sin 3x}{{{D}^{2}}-1} \\ & =x\frac{\sin 3x}{{{D}^{2}}-1}+\frac{d}{dD}\left( \frac{1}{{{D}^{2}}-1} \right)\sin 3x \\ & =x\frac{\sin 3x}{-{{3}^{2}}-1}-{{\left( {{D}^{2}}-1 \right)}^{-2}}\left( 2D \right)\sin 3x \\ & =-\frac{x\sin 3x}{10}-{{\left( {{D}^{2}}-1 \right)}^{-2}}\left( 6\cos 3x \right) \\ & =-\frac{x\sin 3x}{10}-\frac{\left( 6\cos 3x \right)}{{{\left( {{D}^{2}}-1 \right)}^{2}}} \\ & =-\frac{x\sin 3x}{10}-\frac{\left( 6\cos 3x \right)}{100} \\ & =-\frac{1}{10}\left( x\sin 3x+\frac{3\cos 3x}{5} \right) \\ \end{align}\]\[\begin{align} & \left( {{D}^{4}}+2{{D}^{2}}+1 \right)y={{x}^{2}}\cos x \\ & {{y}_{P}}=\frac{{{x}^{2}}\cos x}{\left( {{D}^{4}}+2{{D}^{2}}+1 \right)} \\ & =RP\left( \frac{{{x}^{2}}{{e}^{ix}}}{\left( {{D}^{4}}+2{{D}^{2}}+1 \right)} \right) \\ & =RP\left( {{e}^{ix}}\frac{{{x}^{2}}}{{{\left( {{D}^{2}}+1 \right)}^{2}}} \right) \\ & =RP\left( {{e}^{ix}}\frac{{{x}^{2}}}{{{\left( {{\left( D+i \right)}^{2}}+1 \right)}^{2}}} \right) \\ & =RP\left( {{e}^{ix}}\frac{{{x}^{2}}}{{{\left( \left( {{D}^{2}}-1+2Di \right)+1 \right)}^{2}}} \right) \\ & =RP\left( {{e}^{ix}}\frac{{{x}^{2}}}{{{\left( {{D}^{2}}+2Di \right)}^{2}}} \right) \\ & =RP\left( {{e}^{ix}}\frac{1}{{{\left( D+2i \right)}^{2}}}\frac{{{x}^{2}}}{{{D}^{2}}} \right) \\ & =RP\left( \frac{{{e}^{ix}}}{{{\left( 2i \right)}^{2}}}\frac{1}{\left( 1+\frac{D}{2i} \right)}\frac{1}{\left( 1+\frac{D}{2i} \right)}\frac{{{x}^{4}}}{12} \right) \\ & =RP\left( -\frac{{{e}^{ix}}}{4}\frac{1}{\left( 1+\frac{D}{2i} \right)}\left( 1-\frac{D}{2i}+{{\left( \frac{D}{2i} \right)}^{2}}-{{\left( \frac{D}{2i} \right)}^{3}}+{{\left( \frac{D}{2i} \right)}^{4}} \right)\frac{{{x}^{4}}}{12} \right) \\ & =RP\left( {{e}^{ix}}\frac{1}{\left( 1+\frac{D}{2i} \right)}\left( \frac{{{x}^{4}}}{12}-\frac{{{x}^{3}}}{6i}-\frac{{{x}^{2}}}{4}-\frac{x}{4i}-\frac{1}{8} \right) \right) \\ & =RP\left( {{e}^{ix}}\left( 1-\frac{D}{2i}+{{\left( \frac{D}{2i} \right)}^{2}}-{{\left( \frac{D}{2i} \right)}^{3}}+{{\left( \frac{D}{2i} \right)}^{4}} \right)\left( \frac{{{x}^{4}}}{12}-\frac{{{x}^{3}}}{6i}-\frac{{{x}^{2}}}{4}-\frac{x}{4i}-\frac{1}{8} \right) \right) \\ & =RP\left( {{e}^{ix}}\left( \frac{{{x}^{4}}}{12}-\frac{{{x}^{3}}}{6i}-\frac{{{x}^{2}}}{4}-\frac{x}{4i}-\frac{1}{8}-\left( \frac{{{x}^{3}}}{6i}+\frac{{{x}^{2}}}{4}-\frac{x}{4i}-\frac{1}{8i} \right)+\left( \frac{3{{x}^{2}}}{-12}+\frac{2x}{8i}+\frac{1}{8} \right)+\frac{1}{8i}\left( \frac{6x}{-24i}-\frac{1}{8} \right)+\left( \frac{1}{8} \right) \right) \right) \\ & =i\sin x\left( -\frac{{{x}^{3}}}{8i}-\frac{i}{4}+\frac{x}{4i}+\frac{1}{16i}+\frac{1}{8i}\left( \frac{6}{64} \right) \right) \\ & =\sin x\left( -\frac{{{x}^{3}}}{8}+\frac{1}{4}+\frac{x}{4}+\frac{1}{16}+\frac{3}{256} \right) \\ \end{align}\]
\begin{align*}
&(D^4+2D^2+1)y \;=\; x^2\cos x \\[6pt]
&\text{Note: } \; D^4+2D^2+1 \;=\; (D^2+1)^2. \\[6pt]
&\text{Let } v=(D^2+1)y. \text{ Then } (D^2+1)v = x^2\cos x.
\end{align*}
% ---------- Solve (D^2+1)v = x^2 cos x ----------
\begin{align*}
&\text{Seek } v_p \text{ of the form } v_p = P(x)\cos x + Q(x)\sin x,\\
&\text{and recall the identity } \\
&(D^2+1)\bigl[P\cos x + Q\sin x\bigr]
\;=\;\bigl(P'' + 2Q'\bigr)\cos x + \bigl(Q'' - 2P'\bigr)\sin x.
\end{align*}
\begin{align*}
&\text{To match } x^2\cos x \text{ we need } P''+2Q' \text{ to be degree }2,
\text{ so take } \\
&P(x)=a_3x^3+a_2x^2+a_1x+a_0, \\
&Q(x)=b_3x^3+b_2x^2+b_1x+b_0.
\end{align*}
\begin{align*}
&\text{Compute: } P''=6a_3x+2a_2,\quad Q'=3b_3x^2+2b_2x+b_1,\\
&\quad P''+2Q' = 6b_3x^2+(6a_3+4b_2)x+(2a_2+2b_1),\\
&\quad Q''-2P' = -6a_3x^2+(6b_3-4a_2)x+(2b_2-2a_1).
\end{align*}
\begin{align*}
&\text{Equate with } x^2\cos x + 0\sin x: \\[4pt]
&6b_3 = 1,\\
&6a_3+4b_2 = 0,\\
&2a_2+2b_1 = 0,\\[4pt]
&-6a_3 = 0,\\
&6b_3 -4a_2 = 0,\\
&2b_2 -2a_1 = 0.
\end{align*}
\begin{align*}
&\text{Solve these: } a_3=0,\; b_3=\tfrac{1}{6},\; b_2=0,\; a_2=\tfrac{3}{16},\; b_1=-a_2=-\tfrac{3}{16},\; a_1=0.
\end{align*}
\begin{align*}
&\text{(We may choose the additive constants }a_0,b_0\text{ to be }0\text{ for a particular solution.)}\\[6pt]
&\therefore\quad
P(x)=\tfrac{1}{4}x^2,\qquad
Q(x)=\tfrac{1}{6}x^3-\tfrac{1}{4}x.
\end{align*}
\begin{align*}
&\text{Hence one particular solution for }v\text{ is }\\[4pt]
&\boxed{\; v_p \;=\; \tfrac{1}{4}x^2\cos x \;+\; \Bigl(\tfrac{1}{6}x^3-\tfrac{1}{4}x\Bigr)\sin x \; }.
\end{align*}
% ---------- Now solve (D^2+1) y = v_p ----------
\begin{align*}
&\text{Now solve } (D^2+1)y = v_p. \text{ Seek } y_p = U(x)\cos x + V(x)\sin x,\\
&\text{with } U(x),V(x) \text{ polynomials. Using the same identity: }\\
&(D^2+1)\bigl[U\cos x + V\sin x\bigr] = (U''+2V')\cos x + (V''-2U')\sin x.
\end{align*}
\begin{align*}
&\text{Compare with } v_p \text{ where } \\
&P(x)=\tfrac{1}{4}x^2,\qquad Q(x)=\tfrac{1}{6}x^3-\tfrac{1}{4}x.
\end{align*}
\begin{align*}
&\text{To match degrees we take } U \text{ of degree }4,\; V \text{ of degree }3: \\
&U(x)=u_4x^4+u_3x^3+u_2x^2+u_1x+u_0,\\
&V(x)=v_3x^3+v_2x^2+v_1x+v_0.
\end{align*}
\begin{align*}
&\text{Compute derivatives: } \\
&U''=12u_4x^2+6u_3x+2u_2,\quad V'=3v_3x^2+2v_2x+v_1,\\
&V''=6v_3x+2v_2,\quad U'=4u_4x^3+3u_3x^2+2u_2x+u_1.
\end{align*}
\begin{align*}
&\text{Thus } U''+2V' = (12u_4+6v_3)x^2+(6u_3+4v_2)x+(2u_2+2v_1),\\
&\quad V''-2U' = -8u_4x^3 -6u_3x^2 +(6v_3-4u_2)x + (2v_2-2u_1).
\end{align*}
\begin{align*}
&\text{Equate with } P(x)=\tfrac{1}{4}x^2,\; Q(x)=\tfrac{1}{6}x^3-\tfrac{1}{4}x :\\[4pt]
&12u_4+6v_3 = \tfrac{1}{4}, \tag{I}\\
&6u_3+4v_2 = 0, \tag{II}\\
&2u_2+2v_1 = 0, \tag{III}\\[4pt]
&-8u_4 = \tfrac{1}{6}, \tag{IV}\\
&-6u_3 = 0, \tag{V}\\
&6v_3-4u_2 = -\tfrac{1}{4}, \tag{VI}\\
&2v_2 -2u_1 = 0. \tag{VII}
\end{align*}
\begin{align*}
&\text{Solve these: from (IV) } u_4 = -\tfrac{1}{48}.\\
&\text{From (V) } u_3 = 0.\\
&\text{From (I) } 12(-\tfrac{1}{48}) + 6v_3 = \tfrac{1}{4} \Rightarrow -\tfrac{1}{4}+6v_3=\tfrac{1}{4}
\Rightarrow v_3=\tfrac{1}{12}.\\
&\text{From (VI) } 6(\tfrac{1}{12}) -4u_2 = -\tfrac{1}{4} \Rightarrow \tfrac{1}{2}-4u_2=-\tfrac{1}{4}
\Rightarrow u_2=\tfrac{3}{16}.\\
&\text{From (III) } u_2+v_1=0 \Rightarrow v_1=-\tfrac{3}{16}.\\
&\text{From (II) and (VII) give } v_2=0,\; u_1=0.
\end{align*}
\begin{align*}
&\text{(Choose }u_0=v_0=0\text{ for a particular solution.)}\\[6pt]
&\therefore\quad
U(x) = -\tfrac{1}{48}x^4 + \tfrac{3}{16}x^2, \qquad
V(x) = \tfrac{1}{12}x^3 - \tfrac{3}{16}x.
\end{align*}
\begin{align*}
&\text{So a particular solution of the original equation is }\\[4pt]
&\boxed{\, y_p \;=\; \Bigl(-\tfrac{1}{48}x^4 + \tfrac{3}{16}x^2\Bigr)\cos x
\;+\; \Bigl(\tfrac{1}{12}x^3 - \tfrac{3}{16}x\Bigr)\sin x \, }.
\end{align*}
% ---------- Complementary solution ----------
\begin{align*}
&\text{The homogeneous equation }(D^2+1)^2 y=0\text{ has characteristic roots } r=\pm i \text{ each of multiplicity }2.\\
&\text{Hence the complementary solution is }\\[4pt]
&\boxed{\; y_c \;=\; \bigl(C_1 + C_2 x\bigr)\cos x \;+\; \bigl(C_3 + C_4 x\bigr)\sin x \; }.
\end{align*}
\begin{align*}
&\text{Therefore the general solution is}\\[4pt]
&\boxed{\; y(x) \;=\; \bigl(C_1 + C_2 x\bigr)\cos x + \bigl(C_3 + C_4 x\bigr)\sin x \;+\;
\Bigl(-\tfrac{1}{48}x^4 + \tfrac{3}{16}x^2\Bigr)\cos x
+\Bigl(\tfrac{1}{12}x^3 - \tfrac{3}{16}x\Bigr)\sin x \; }.
\end{align*}
\(\left( {{D}^{2}}+1 \right)y=x{{e}^{2x}}\)
Method-1
\[\begin{align} & \left( {{D}^{2}}+1 \right)y=x{{e}^{2x}} \\ & \because {{y}_{PI}}=\frac{xf\left( x \right)}{f\left( D \right)}=x\frac{f\left( x \right)}{f\left( D \right)}+\frac{d}{dD}\left( \frac{1}{f\left( D \right)} \right)f\left( x \right) \\ \Rightarrow {{y}_{PI}}&=\frac{x{{e}^{2x}}}{{{D}^{2}}+1} \\ & =x\frac{{{e}^{2x}}}{{{D}^{2}}+1}+\frac{d}{dD}\left( \frac{1}{{{D}^{2}}+1} \right){{e}^{2x}} \\ & =x\frac{{{e}^{2x}}}{{{2}^{2}}+1}+\left( \frac{-2D}{{{\left( {{2}^{2}}+1 \right)}^{2}}} \right){{e}^{2x}} \\ & =x\frac{{{e}^{2x}}}{5}+\left( \frac{-4}{25} \right){{e}^{2x}} \\ & =\frac{{{e}^{2x}}}{25}\left( 5x-4 \right) \\ \end{align}\]Method-2
\[\begin{align} \left( {{D}^{2}}+1 \right)y& =x{{e}^{2x}} \\ & \Rightarrow {{y}_{PI}}=\frac{x{{e}^{2x}}}{{{D}^{2}}+1} \\ & ={{e}^{2x}}\frac{1}{{{D}^{2}}+1}\left( x \right) \\ & ={{e}^{2x}}\frac{1}{{{\left( D+2 \right)}^{2}}+1}\left( x \right) \\ & ={{e}^{2x}}\frac{1}{{{D}^{2}}+4D+5}\left( x \right) \\ & =\frac{{{e}^{2x}}}{5}\frac{1}{\left( 1+\frac{{{D}^{2}}+4D}{5} \right)}\left( x \right) \\ & =\frac{{{e}^{2x}}}{5}{{\left( 1+\left( \frac{{{D}^{2}}+4D}{5} \right) \right)}^{-1}}\left( x \right) \\ & =\frac{{{e}^{2x}}}{5}\left( 1-\left( \frac{{{D}^{2}}+4D}{5} \right)+{{\left( \frac{{{D}^{2}}+4D}{5} \right)}^{2}}+.... \right)\left( x \right) \\ & =\frac{{{e}^{2x}}}{5}\left( x-\frac{4}{5} \right) \\ & =\frac{{{e}^{2x}}}{25}\left( 5x-4 \right) \\ \end{align}\]\(\left( {{D}^{2}}-2D+4 \right)y={{e}^{x}}\cos x\)
Method-1
\[\begin{align} & \left( {{D}^{2}}-2D+4 \right)y={{e}^{x}}\cos x \\ & {{y}_{PI}}=\frac{{{e}^{x}}\cos x}{{{D}^{2}}-2D+4} \\ & ={{e}^{x}}\frac{1}{{{\left( D+1 \right)}^{2}}-2\left( D+1 \right)+4}\left( \cos x \right) \\ & ={{e}^{x}}\frac{1}{{{D}^{2}}+3}\left( \cos x \right) \\ & ={{e}^{x}}\frac{1}{-{{1}^{2}}+3}\left( \cos x \right) \\ & =\frac{{{e}^{x}}\cos x}{2} \\ \end{align}\]Method-2
\[\begin{align} & \left( {{D}^{2}}-2D+4 \right)y={{e}^{x}}\cos x \\ {{y}_{PI}}&=\frac{{{e}^{x}}\cos x}{{{D}^{2}}-2D+4} \\ & =RP\left( \frac{{{e}^{x}}{{e}^{ix}}}{{{D}^{2}}-2D+4} \right) \\ & =RP\left( \frac{{{e}^{\left( 1+i \right)x}}}{{{D}^{2}}-2D+4} \right) \\ & =RP\left( \frac{{{e}^{\left( 1+i \right)x}}}{{{\left( 1+i \right)}^{2}}-2\left( 1+i \right)+4} \right) \\ & =RP\left( \frac{{{e}^{\left( 1+i \right)x}}}{1-1+2i-2-2i+4} \right) \\ & =RP\left( \frac{{{e}^{\left( 1+i \right)x}}}{2} \right) \\ & ={{e}^{x}}RP\left( \frac{\cos x+i\sin x}{2} \right) \\ & =\frac{{{e}^{x}}\cos x}{2} \\ \end{align}\]\(\left( {{D}^{2}}-2D+1 \right)y=x{{e}^{x}}\sin x \)
Method-1
\[\begin{align} & \left( {{D}^{2}}-2D+1 \right)y=x{{e}^{x}}\sin x \\ {{y}_{PI}} &=x\frac{{{e}^{x}}\sin x}{\left( {{D}^{2}}-2D+1 \right)}+\frac{1}{dD}\left( \frac{1}{{{D}^{2}}-2D+1} \right){{e}^{x}}\sin x \\ & =x{{e}^{x}}\frac{\sin x}{{{\left( D-1 \right)}^{2}}}+\frac{1}{dD}\left( \frac{1}{{{\left( D-1 \right)}^{2}}} \right){{e}^{x}}\sin x \\ & =x{{e}^{x}}\frac{\sin x}{{{\left( D+1-1 \right)}^{2}}}+\left( \frac{-2}{{{\left( D-1 \right)}^{3}}} \right){{e}^{x}}\sin x \\ & =x{{e}^{x}}\frac{\sin x}{{{D}^{2}}}-2{{e}^{x}}\left( \frac{1}{{{\left( D+1-1 \right)}^{3}}} \right)\sin x \\ & =x{{e}^{x}}\frac{\sin x}{-{{1}^{2}}}-2{{e}^{x}}\left( \frac{1}{{{D}^{3}}} \right)\sin x \\ & =x{{e}^{x}}\frac{\sin x}{-{{1}^{2}}}-2{{e}^{x}}\left( \frac{1}{D{{D}^{2}}} \right)\sin x \\ & =-x{{e}^{x}}\sin x-2{{e}^{x}}\frac{1}{{{D}^{2}}}\frac{1}{D}\sin x \\ & =-x{{e}^{x}}\sin x-2{{e}^{x}}\left( \frac{-\cos x}{{{D}^{2}}} \right) \\ & =-x{{e}^{x}}\sin x+2{{e}^{x}}\left( \frac{\cos x}{-{{1}^{2}}} \right) \\ & =-x{{e}^{x}}\sin x-2{{e}^{x}}\cos x \\ & =-{{e}^{x}}\left( 2\cos x+x\sin x \right) \\ \end{align}\]Method-2
\[\begin{align} & \left( {{D}^{2}}-2D+1 \right)y=x{{e}^{x}}\sin x \\ {{y}_{PI}} &=IP\left( \frac{x{{e}^{x}}{{e}^{ix}}}{\left( {{D}^{2}}-2D+1 \right)} \right) \\ & =IP\left( \frac{x{{e}^{\left( 1+i \right)x}}}{{{\left( D-1 \right)}^{2}}} \right) \\ & =IP\left( {{e}^{\left( 1+i \right)x}}\frac{x}{{{\left( D+1+i-1 \right)}^{2}}} \right) \\ & =IP\left( {{e}^{\left( 1+i \right)x}}\frac{x}{{{\left( D+i \right)}^{2}}} \right) \\ & =IP\left( {{e}^{\left( 1+i \right)x}}\frac{1}{{{i}^{2}}}\frac{x}{{{\left( 1-Di \right)}^{2}}} \right) \\ & =-IP\left( {{e}^{\left( 1+i \right)x}}\left( 1+2Di+3{{D}^{2}}{{i}^{2}}+... \right)x \right) \\ & =-IP\left( {{e}^{\left( 1+i \right)x}}\left( x+2i \right) \right) \\ & =-{{e}^{x}}IP\left[ \left( \cos x+i\sin x \right)\left( x+2i \right) \right] \\ & =-{{e}^{x}}\left( 2\cos x+x\sin x \right) \\ \end{align}\]Solutions of the differential equations
\( \left( {{D}^{2}}+5D+6 \right)y ={{e}^{x}} \)
Solutions
The Complementary function
The auxiliary equations is : \begin{align*} & {{m}^{2}}+5m+6 =0 \\ & \Rightarrow \left( m+3 \right)\left( m+2 \right)=0 \\ \text{Thus the roots are} m=-3,m=-2 \\ & \Rightarrow {{y}_{CF}}={{c}_{1}}{{e}^{-3x}}+{{c}_{2}}{{e}^{-2x}} \\ \end{align*}The particular Integral is :
\[\begin{align} {{y}_{PI}} &=\frac{{{e}^{x}}}{{{D}^{2}}+5D+6} \\ & =\frac{{{e}^{x}}}{{{1}^{2}}+5\left( 1 \right)+6} \\ & =\frac{{{e}^{x}}}{12} \\ \end{align}\]The Solution of the differential equation is :
\[\begin{align} y&={{y}_{CF}}+{{y}_{PI}} \\ & ={{c}_{1}}{{e}^{-3x}}+{{c}_{2}}{{e}^{-2x}}+\frac{{{e}^{x}}}{12}\, \\ \end{align}\]\( \left( {{D}^{2}}+1 \right)y =\cos \left( 2x-1 \right) \)
\[\begin{align} {{m}^{2}}+1 &=0 \\ & \Rightarrow m=\pm i \\ & \Rightarrow {{y}_{CF}}=A\cos x+B\sin x \\ \end{align}\] \[\begin{align} {{y}_{PI}}&=\frac{\cos \left( 2x-1 \right)}{{{D}^{2}}+1} \\ & =\frac{\cos \left( 2x-1 \right)}{-{{2}^{2}}+1} \\ & =-\frac{1}{3}\cos \left( 2x-1 \right) \\ \end{align}\] The solution is : \[\begin{align} y&={{y}_{CF}}+{{y}_{PI}} \\ & \therefore y=A\cos x+B\sin x-\frac{1}{3}\cos \left( 2x-1 \right) \\ \end{align}\]\( \left( {{D}^{3}}+4D \right)y=\sin 2x \)
\[\begin{align} & {{m}^{3}}+4m =0 \\ & \Rightarrow m\left( {{m}^{2}}+4 \right)=0 \\ & \Rightarrow m=0,m=2i,-2i \\ & \Rightarrow {{y}_{CF}}={{c}_{1}}{{e}^{0x}}+A\cos 2x+B\sin 2x \\ & \Rightarrow {{y}_{CF}}=A\cos 2x+B\sin 2x+c \\ \end{align}\] \[\begin{align} {{y}_{PI}}&=\frac{\sin 2x}{{{D}^{3}}+4D} \\ & =\frac{\sin 2x}{D\left( {{D}^{2}}+4 \right)} \\ & =x\frac{\sin 2x}{{{f}^{'}}\left( -{{a}^{2}} \right)} \\ & =x\frac{\sin 2x}{{{\left( 3{{D}^{2}}+4 \right)}_{{{D}^{2}}=-{{a}^{2}}}}} \\ & =x\frac{\sin 2x}{\left( 3\left( -{{2}^{2}} \right)+4 \right)} \\ & =x\frac{\sin 2x}{\left( -12+4 \right)}\, \\ & =-\frac{1}{8}x\sin 2x \\ \end{align}\]\( \left( {{D}^{2}}+D \right)y={{x}^{2}}+2x+4 \)
\[\begin{align} & {{m}^{2}}+m=0 \\ & \Rightarrow m\left( m+1 \right)=0 \\ & \Rightarrow \,m=0,m=-1 \\ & \Rightarrow {{y}_{Cf}}={{c}_{1}}{{e}^{0x}}+{{c}_{2}}{{e}^{-x}} \\ & \Rightarrow {{y}_{Cf}}={{c}_{1}}+{{c}_{2}}{{e}^{-x}} \\ \end{align}\] \[\begin{align} {{y}_{Pi}}& =\frac{{{x}^{2}}+2x+4}{{{D}^{2}}+D} \\ & =\frac{{{x}^{2}}+2x+4}{D\left( D+1 \right)} \\ & =\frac{1}{1+D}\frac{1}{D}\left( {{x}^{2}}+2x+4 \right) \\ & =\frac{1}{1+D}\int{\left( {{x}^{2}}+2x+4 \right)}\,dx \\ & =\left( 1-D+{{D}^{2}}-{{D}^{3}}+.... \right)\left( \frac{{{x}^{3}}}{3}+{{x}^{2}}+4x \right) \\ & =\left( \frac{{{x}^{3}}}{3}+{{x}^{2}}+4x \right)-D\left( \frac{{{x}^{3}}}{3}+{{x}^{2}}+4x \right)+{{D}^{2}}\left( \frac{{{x}^{3}}}{3}+{{x}^{2}}+4x \right)-{{D}^{3}}\left( \frac{{{x}^{3}}}{3}+{{x}^{2}}+4x \right) \\ & =\left( \frac{{{x}^{3}}}{3}+{{x}^{2}}+4x \right)-\left( {{x}^{2}}+2x+4 \right)+\left( 2x+2 \right)-2 \\ & =\frac{{{x}^{3}}}{3}+4x-4 \\ \end{align}\] \[y={{y}_{Cf}}+{{y}_{Pi}}={{c}_{1}}+{{c}_{2}}{{e}^{-x}}+\frac{{{x}^{3}}}{3}+4x-4\]\(\left( {{D}^{2}}-2D+4 \right)y={{e}^{x}}\cos x \)
\[\begin{align} & {{m}^{2}}-2m+4=0 \\ & \Rightarrow m=\frac{2\pm \sqrt{4-16}}{2} \\ & =\frac{2\pm \sqrt{-12}}{2}=1\pm \sqrt{3}i \\ & \Rightarrow {{y}_{Cf}}={{e}^{x}}\left( A\sin \sqrt{3}x+B\cos \sqrt{3}x \right) \\ & {{y}_{Pi}}=\frac{{{e}^{x}}\cos x}{{{D}^{2}}-2D+4} \\ & ={{e}^{x}}\frac{\cos x}{f\left( D+a \right)} \\ & ={{e}^{x}}\frac{\cos x}{{{\left( D+1 \right)}^{2}}-2\left( D+1 \right)+4} \\ & ={{e}^{x}}\frac{\cos x}{{{D}^{2}}+1+2D-2D-2+4} \\ & ={{e}^{x}}\frac{\cos x}{{{D}^{2}}+3} \\ & ={{e}^{x}}\frac{\cos x}{-{{1}^{2}}+3} \\ & ={{e}^{x}}\frac{\cos x}{2} \\ & \Rightarrow y={{y}_{Cf}}+{{y}_{Pi}} \\ & \Rightarrow y={{e}^{x}}\left( A\sin \sqrt{3}x+B\cos \sqrt{3}x \right)+\frac{{{e}^{x}}}{2}\cos x \\ \end{align}\]\(\left( {{D}^{2}}-2D+4 \right)y=x\cos x \)
\[\begin{align} {{y}_{PI}} &=\frac{x\cos x}{{{D}^{2}}-2D+4} \\ & =\text{Imaginary Part of }\frac{x{{e}^{ix}}}{{{D}^{2}}-2D+4} \\ & ={{e}^{ix}}\frac{x}{{{\left( D+i \right)}^{2}}-2\left( D+i \right)+4} \\ & ={{e}^{ix}}\frac{x}{{{D}^{2}}-1+2iD-2D-2i+4} \\ & ={{e}^{ix}}\frac{x}{{{D}^{2}}+2iD-2D-2i+3} \\ & ={{e}^{ix}}\frac{x}{{{D}^{2}}-2D+3+2iD-2i} \\ & ={{e}^{ix}}\frac{x}{{{D}^{2}}-2D+3+2i(D-1)} \\ & =\frac{{{e}^{ix}}}{2i(D-1)}\left( \frac{x}{1+\frac{{{D}^{2}}-2D+3}{2i(D-1)}} \right) \\ & =\frac{{{e}^{ix}}}{-2i}\left( \frac{1}{1+\frac{{{D}^{2}}-2D+3}{2i(D-1)}} \right)\frac{x}{1-D} \\ & =\frac{{{e}^{ix}}}{-2i}\left( \frac{1}{1+\frac{{{D}^{2}}-2D+3}{2i(D-1)}} \right)\left( 1+D+{{D}^{2}}+... \right)\left( x \right) \\ & =\frac{{{e}^{ix}}}{-2i}\left( \frac{1}{1+\frac{{{D}^{2}}-2D+3}{2i(D-1)}} \right)\left( x+1 \right) \\ & =\frac{{{e}^{ix}}}{-2i}\left( 1-\left( \frac{{{D}^{2}}-2D+3}{2i(D-1)} \right)+{{\left( \frac{{{D}^{2}}-2D+3}{2i(D-1)} \right)}^{2}}+... \right)\left( x+1 \right) \\ & =\frac{{{e}^{ix}}}{-2i}\left( x+1-\frac{-2+3x+3}{2i(D-1)} \right) \\ & =\frac{{{e}^{ix}}}{-2i}\left( x+1-\frac{3x+1}{2i(D-1)} \right) \\ & =\frac{{{e}^{ix}}}{-2i}\left( x+1+\frac{3x+1}{2i(1-D)} \right) \\ & =\frac{{{e}^{ix}}}{-2i}\left( x+1+\frac{1}{2i}\left( 1+D+{{D}^{2}}+... \right)\left( 3x+1 \right) \right) \\ & =\frac{{{e}^{ix}}}{-2i}\left( x+1+\frac{1}{2i}\left( 3x+1+3 \right) \right) \\ & =\frac{i{{e}^{ix}}}{2}\left( x+1+\frac{1}{2i}\left( 3x+1+3 \right) \right)\, \\ & =\frac{{{e}^{ix}}}{2}\left( i\left( x+1 \right)+\frac{1}{2}\left( 3x+4 \right) \right)\, \\ & =\frac{\cos x+i\sin x}{2}\left( i\left( x+1 \right)+\frac{1}{2}\left( 3x+4 \right) \right) \\ & =\frac{\cos x\left( 3x+4 \right)}{4}-\frac{\sin x\left( 3x+4 \right)}{2}+i\left( \frac{\cos x\left( x+1 \right)}{2}+\frac{\sin x\left( 3x+4 \right)}{4} \right) \\ & \text{Imaginary Part of }{{y}_{PI}}=\frac{\cos x\left( x+1 \right)}{2}+\frac{\sin x\left( 3x+4 \right)}{4}\, \\ \end{align}\]\( \left( {{D}^{2}}-2D+1 \right)y=x{{e}^{x}}\sin x\)
\[\begin{align} & {{m}^{2}}-2m+1=0 \\ & \Rightarrow {{\left( m-1 \right)}^{2}}=0 \\ & \Rightarrow m=1,1 \\ & \Rightarrow {{y}_{Cf}}=\left( {{c}_{1}}+{{c}_{2}}x \right){{e}^{x}} \\ {{y}_{Pi}}& =\frac{x{{e}^{x}}\sin x}{{{D}^{2}}-2D+1} \\ & ={{e}^{x}}\frac{x\sin x}{f\left( D+a \right)} \\ & ={{e}^{x}}\frac{x\sin x}{{{\left( D+1-1 \right)}^{2}}} \\ & ={{e}^{x}}\frac{x\sin x}{{{\left( D+1-1 \right)}^{2}}} \\ & ={{e}^{x}}\frac{x\sin x}{{{\left( D \right)}^{2}}} \\ & ={{e}^{x}}\frac{1}{D}\int{x\sin xdx} \\ & ={{e}^{x}}\frac{1}{D}\left( -x\cos x+\sin x \right) \\ & ={{e}^{x}}\int{\left( -x\cos x+\sin x \right)}dx \\ & ={{e}^{x}}\left( -x\sin x-\cos x-\cos x \right) \\ & =-{{e}^{x}}\left( x\sin x+2\cos x \right) \\ & \Rightarrow y={{y}_{CF}}+{{y}_{PI}}=\left( {{c}_{1}}+{{c}_{2}}x \right){{e}^{x}}-{{e}^{x}}\left( x\sin x+2\cos x \right) \\ & \Rightarrow y={{e}^{x}}\left( {{c}_{1}}+{{c}_{2}}x-x\sin x-2\cos x \right) \end{align}\]\( \left( {{D}^{2}}-4 \right)y=x\sinh x \)
\[\begin{align} & {{m}^{2}}-4=0 \\ & \Rightarrow m=\pm 2 \\ \Rightarrow {{y}_{CF}} &={{c}_{1}}{{e}^{2x}}+{{c}_{2}}{{e}^{-2x}} \\ \Rightarrow {{y}_{PI}}&=\frac{x\sinh x}{{{D}^{2}}-4} \\ & =\frac{x\left( \frac{{{e}^{x}}-{{e}^{-x}}}{2} \right)}{{{D}^{2}}-4}=\frac{x\left( {{e}^{x}}-{{e}^{-x}} \right)}{2\left( {{D}^{2}}-4 \right)} \\ & ={{e}^{x}}\frac{x}{2\left( {{\left( D+1 \right)}^{2}}-4 \right)}-{{e}^{-x}}\frac{x}{2\left( {{\left( D-1 \right)}^{2}}-4 \right)} \\ & ={{e}^{x}}\frac{x}{2\left( {{D}^{2}}+2D-3 \right)}-{{e}^{-x}}\frac{x}{2\left( {{D}^{2}}-2D-3 \right)} \\ & =\frac{{{e}^{x}}}{2}\frac{1}{-3}\left( \frac{1}{1-\frac{{{D}^{2}}+2D}{3}} \right)\left( x \right)-\frac{{{e}^{-x}}}{2}\frac{1}{-3}\left( \frac{1}{1-\frac{{{D}^{2}}-2D}{3}} \right)\left( x \right) \\ & =\frac{{{e}^{x}}}{-6}\left( 1+\left( \frac{{{D}^{2}}+2D}{3} \right) \right)\left( x \right)+\frac{{{e}^{-x}}}{6}\left( 1+\left( \frac{{{D}^{2}}-2D}{3} \right) \right)\left( x \right) \\ & =\frac{{{e}^{x}}}{-6}\left( x+\frac{2}{3} \right)+\frac{{{e}^{-x}}}{6}\left( x-\frac{2}{3} \right) \\ & =-\frac{x}{6}\left( {{e}^{x}}-{{e}^{-x}} \right)-\frac{1}{9}\left( {{e}^{x}}+{{e}^{-x}} \right) \\ & =-\frac{x}{3}\sinh x-\frac{2}{9}\cosh x \\ & \Rightarrow y={{y}_{Cf}}+{{y}_{Pi}} \\ & \Rightarrow ={{c}_{1}}{{e}^{2x}}+{{c}_{2}}{{e}^{-2x}}-\frac{x}{3}\sinh x-\frac{2}{9}\cosh x \\ \end{align}\]
Let us take an example and solve by the method-1 called Inverse Operator Method
Example: solve: \( \frac{{{d}^{3}}y}{d{{x}^{3}}}+2y=\sin 3x \)
\[\begin{align} & \frac{{{d}^{3}}y}{d{{x}^{3}}}+2y=\sin 3x \\ & \left( {{D}^{3}}+2 \right)y=\sin 3x \\ y_(PI)&=\frac{\sin 3x}{{{D}^{3}}+2} \\ & =\frac{\sin x}{D{{D}^{2}}+2} \\ & =\frac{\sin x}{D\left( -{{3}^{2}} \right)+2} \\ & =\frac{\left( -9D-2 \right)\sin 3x}{81{{D}^{2}}-4} \\ & =\frac{-27\cos3 x-2\sin 3x}{81\left( -{{3}^{2}} \right)-4} \\ & =\frac{-27\cos3 x-2\sin 3x}{-729-4} \\ & =\frac{27\cos 3x+2\sin 3x}{731} \\ \end{align}\]
Let us compare by the method-2 called the method of variation of parameters
\[\begin{align}
{{y}_{P}}&=A\sin 3x+B\cos 3x \\
D{{y}_{P}}&=-3A\cos 3x-3B\sin 3x \\
{{D}^{2}}{{y}_{P}} &=-9A\sin 3x-9B\cos 3x \\
{{D}^{3}}{{y}_{P}} &=27A\cos 3x+27B\sin 3x \\
& \Rightarrow \left( {{D}^{3}}+2 \right){{y}_{P}}=\sin 3x \\
& \Rightarrow 27A\cos 3x-27B\sin 3x+2A\sin 3x+2B\cos 3x=\sin 3x \\
& \Rightarrow \left( 27A+2B \right)\cos 3x+\left( 2A-27B \right)\sin 3x=\sin 3x \\
& \Rightarrow 27A+2B=0\And 2A-27B=1 \\
& \Rightarrow 27A+2\frac{\left( 2A-1 \right)}{27\,}\,=0 \\
& \Rightarrow {{27}^{2}}A+4A=2 \\
& \Rightarrow A\left( 731 \right)=2 \\
& \Rightarrow A=\frac{2}{731},B=-\frac{27A}{2}=-\frac{27}{731} \\
& \Rightarrow {{y}_{P}}=\frac{2}{731}\sin 3x+\frac{27}{731}\cos 3x \\
\end{align}\]
Solve \( \frac{dy}{dx}=\sin 2x \)
Method 1 \[\begin{align} & \frac{dy}{dx}=\sin 2x \\ & \Rightarrow Dy=\sin 2x \\ \Rightarrow {{y}_{PI}}& =\frac{\sin 2x}{D} \\ & =\int{\sin 2xdx} \\ & =-\frac{1}{2}\cos 2x \\ \end{align}\]
Method2
\[\begin{align}
& \frac{dy}{dx}=\sin 2x \\
& \Rightarrow Dy=\sin 2x \\
\Rightarrow {{y}_{PI}}&=\frac{\sin 2x}{D} \\
& =\frac{D\left( \sin 2x \right)}{{{D}^{2}}} \\
& =\frac{2\cos 2x}{-{{2}^{2}}} \\
& =-\frac{1}{2}\cos 2x \\
\end{align}\]
Example \( (D^2 - 3D + 2)y = e^{2x} \)
Method-1(Inverse Operator Method)
\[\begin{align} {{y}_{PI}}&=\frac{{{e}^{2x}}}{{{D}^{2}}-3D+2} \\ & =x\frac{{{e}^{2x}}}{{{f}^{'}}\left( a \right)} \\ & =x\frac{{{e}^{2x}}}{{{\left( 2D-3 \right)}_{D=2}}} \\ & =x\frac{{{e}^{2x}}}{2\times 2-3} \\ & =x{{e}^{2x}} \\ \end{align}\]Method-2 (Method of Undetermined Coefficients)
Example \( (D^2 - 3D + 2)y = e^{2x} \)
Step 1: CF
\[ m^2 - 3m + 2 = 0 \Rightarrow m = 1, 2 \] \[ y_c = C_1 e^x + C_2 e^{2x} \]Step 2: PI
RHS: \(e^{2x}\), resonance → multiply by x:
\[ y_p = A x e^{2x} \]Step 3: Derivatives & substitution
\[ y_p' = A e^{2x} + 2 A x e^{2x}, \quad y_p'' = 4 A x e^{2x} + 4 A e^{2x} \] \[ (D^2 - 3D + 2)y_p = A e^{2x} = e^{2x} \Rightarrow A = 1 \]Step 4: General solution
\[ y = C_1 e^x + C_2 e^{2x} + x e^{2x} \]Method-3 ( Variation of Parameters)
The equation to be solved is: $$ (D^2 - 3D + 2)y = e^{2x} $$
The method of variation of parameters is a general technique to find the particular solution when the complementary solution is known.
Step 1: Find the Complementary Solution (\(y_c\))First, solve the homogeneous equation \((D^2 - 3D + 2)y = 0\). The characteristic equation is \(m^2 - 3m + 2 = 0\), which factors to \((m-1)(m-2)=0\). The roots are \(m_1 = 1\) and \(m_2 = 2\).
$$ y_c = c_1 e^x + c_2 e^{2x} $$This gives us the two linearly independent solutions: \(y_1 = e^x\) and \(y_2 = e^{2x}\).
Step 2: Calculate the Wronskian (\(W\))The Wronskian of the solutions \(y_1\) and \( y_2\) is a determinant used in the formula.
$$ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} e^x & e^{2x} \\ e^x & 2e^{2x} \end{vmatrix} $$ $$ W = (e^x)(2e^{2x}) - (e^{2x})(e^x) = 2e^{3x} - e^{3x} = e^{3x} $$ Step 3: Find the Particular Solution (\(y_p\))The particular solution is given by the formula: $$ y_p = -y_1 \int \frac{y_2 f(x)}{W} dx + y_2 \int \frac{y_1 f(x)}{W} dx $$
Here, the non-homogeneous term is \(f(x) = e^{2x}\).
\text{First integral: } -e^x \int \frac{(e^{2x})(e^{2x})}{e^{3x}} dx = -e^x \int e^x dx = -e^x(e^x) = -e^{2x} \] \[ \text{Second integral: } +e^{2x} \int \frac{(e^x)(e^{2x})}{e^{3x}} dx = +e^{2x} \int 1 dx = e^{2x}(x) = xe^{2x} \]Combining the results gives the particular solution:
\[ y_p = -e^{2x} + xe^{2x} \] Step 4: Form the General SolutionThe general solution is the sum of the complementary and particular solutions, \(y = y_c + y_p\).
\[ y = (c_1e^x + c_2e^{2x}) + (-e^{2x} + xe^{2x})\]The term \(-e^{2x}\) can be absorbed into the arbitrary constant \(c_2e^{2x}\), as it is a multiple of a homogeneous solution. Thus, we can write the final solution as:
\[ y = c_1e^x + c_2e^{2x} + xe^{2x} \]Example 2: \( (D^2 + 1)y = \cos x \)
Step 1: CF
\[ m^2 + 1 = 0 \Rightarrow m = \pm i \] \[ y_c = C_1 \cos x + C_2 \sin x \]Step 2: PI
\[ y_p = A x \cos x + B x \sin x \quad (\text{resonance}) \]Step 3: Derivatives & substitution
\[ y_p' = A \cos x - A x \sin x + B \sin x + B x \cos x \] \[ y_p'' = -2 A \sin x - A x \cos x + 2 B \cos x - B x \sin x \] \[ (D^2 + 1)y_p = -2A \sin x + 2B \cos x = \cos x \Rightarrow A = 0, B = \frac{1}{2} \]Step 4: General solution
\[ y = C_1 \cos x + C_2 \sin x + \frac{1}{2} x \sin x \]Example 3: \( (D^2 - 4)y = x \)
Step 1: CF
\[ m^2 - 4 = 0 \Rightarrow m = \pm 2 \] \[ y_c = C_1 e^{2x} + C_2 e^{-2x} \]Step 2: PI
\[ y_p = A x + B \]Step 3: Derivatives & substitution
\[ y_p' = A, \quad y_p'' = 0 \] \[ (D^2 - 4)y_p = -4(Ax + B) = x \Rightarrow A = -\frac{1}{4}, B=0 \]Step 4: General solution
\[ y = C_1 e^{2x} + C_2 e^{-2x} - \frac{1}{4} x \]Example 4: \( (D^2 - 1)y = e^x \)
Step 1: CF
\[ m^2 -1 =0 \Rightarrow m = \pm 1 \] \[ y_c = C_1 e^x + C_2 e^{-x} \]Step 2: PI
\[ y_p = A x e^x \quad (\text{resonance}) \]Step 3: Derivatives & substitution
\[ y_p' = A e^x + A x e^x, \quad y_p'' = 2A e^x + A x e^x \] \[ (D^2 - 1)y_p = A e^x = e^x \Rightarrow A = 1 \]Step 4: General solution
\[ y = C_1 e^x + C_2 e^{-x} + x e^x \]Example 5: \( (D^2 + 4)y = \sin 2x \)
Step 1: CF
\[ m^2 + 4 = 0 \Rightarrow m = \pm 2i \] \[ y_c = C_1 \cos 2x + C_2 \sin 2x \]Step 2: PI
\[ y_p = A x \cos 2x + B x \sin 2x \quad (\text{resonance}) \]Step 3: Derivatives & substitution
\[ (D^2 + 4)y_p = -4A \sin 2x + 4B \cos 2x = \sin 2x \Rightarrow A=-\frac{1}{4}, B=0 \]Step 4: General solution
\[ y = C_1 \cos 2x + C_2 \sin 2x - \frac{1}{4} x \sin 2x \]Example 6: \( (D^2 - 2D + 1)y = e^x \)
Step 1: CF
\[ m^2 - 2m + 1 = 0 \Rightarrow m = 1 \text{ (double root)} \] \[ y_c = (C_1 + C_2 x) e^x \]Step 2: PI
RHS: \(e^x\), \(f(1)=0\), double root → multiply by x²:
\[ y_p = A x^2 e^x \]Step 3: Derivatives & substitution
\[ y_p' = 2A x e^x + A x^2 e^x, \quad y_p'' = 2A e^x + 4A x e^x + A x^2 e^x \] \[ (D^2-2D+1)y_p = 2A e^x = e^x \Rightarrow A = 1/2 \]Step 4: General solution
\[ y = (C_1 + C_2 x) e^x + \frac{1}{2} x^2 e^x \]Example 7: \( (D^2 + D)y = x \)
Step 1: CF
\[ m^2 + m = 0 \Rightarrow m = 0, -1 \] \[ y_c = C_1 + C_2 e^{-x} \]Step 2: PI
RHS: polynomial \(x\), trial: \(y_p = A x + B x^2\)
Step 3: Derivatives & substitution
\[ y_p' = A + 2 B x, \quad y_p'' = 2 B \] \[ (D^2 + D)y_p = 2B + A + 2B x = x \Rightarrow 2B =1, 2B + A=0 \Rightarrow B=1/2, A=-1 \]Step 4: General solution
\[ y = C_1 + C_2 e^{-x} - x + \frac{1}{2} x^2 \]Example 8: \( (D^2 - 5D + 6)y = e^{2x} \)
Step 1: CF
\[ m^2 -5m +6=0 \Rightarrow m=2,3 \] \[ y_c = C_1 e^{2x} + C_2 e^{3x} \]Step 2: PI
RHS: \(e^{2x}\), \(f(2)=0\), resonance → multiply by x:
\[ y_p = A x e^{2x} \]Step 3: Derivatives & substitution
\[ y_p' = A e^{2x} + 2 A x e^{2x}, \quad y_p'' = 4 A x e^{2x} + 4 A e^{2x} \] \[ (D^2-5D+6)y_p = A e^{2x} = e^{2x} \Rightarrow A=1 \]Step 4: General solution
\[ y = C_1 e^{2x} + C_2 e^{3x} + x e^{2x} \]Example 9: \( (D^2 + 4D + 4)y = e^{-2x} \)
Step 1: CF
\[ m^2 +4m +4=0 \Rightarrow m=-2 \text{ (double root)} \] \[ y_c = (C_1 + C_2 x) e^{-2x} \]Step 2: PI
RHS: \(e^{-2x}\), \(f(-2)=0\), double root → multiply by x²:
\[ y_p = A x^2 e^{-2x} \]Step 3: Derivatives & substitution
\[ y_p' = 2A x e^{-2x} -2A x^2 e^{-2x}, \quad y_p'' = 2A e^{-2x} -8A x e^{-2x} +4A x^2 e^{-2x} \] \[ (D^2+4D+4)y_p = 2A e^{-2x} = e^{-2x} \Rightarrow A=1/2 \]Step 4: General solution
\[ y = (C_1 + C_2 x) e^{-2x} + \frac{1}{2} x^2 e^{-2x} \]Example 10: \( (D^2 - D - 2)y = \sin x \)
Step 1: CF
\[ m^2 - m - 2 = 0 \Rightarrow m = 2, -1 \] \[ y_c = C_1 e^{2x} + C_2 e^{-x} \]Step 2: PI
RHS: \(\sin x\), try:
\[ y_p = A \sin x + B \cos x \]Step 3: Derivatives
\[ y_p' = A \cos x - B \sin x, \quad y_p'' = -A \sin x - B \cos x \]Step 4: Apply operator
\[ (D^2 - D - 2)y_p = y_p'' - y_p' - 2y_p = (-A \sin x - B \cos x) - (A \cos x - B \sin x) - 2(A \sin x + B \cos x) \]Combine terms:
\[ \text{sin x term: } -3A + B, \quad \text{cos x term: } -A - 3B \]Step 5: Equate to RHS
\[ -3A + B = 1, \quad -A - 3B = 0 \Rightarrow B = \frac{1}{10}, \quad A = -\frac{3}{10} \]Step 6: Particular Integral
\[ y_p = -\frac{3}{10} \sin x + \frac{1}{10} \cos x \]Step 7: General solution
\[ y = C_1 e^{2x} + C_2 e^{-x} - \frac{3}{10} \sin x + \frac{1}{10} \cos x \]Example 1: \( (D^2 - 1)y = x e^x \)
Step 1: CF
\[ m^2 -1 =0 \Rightarrow m = \pm 1 \quad y_c = C_1 e^x + C_2 e^{-x} \]Step 2: PI
RHS = x e^x, trial: \(y_p = (Ax^2 + Bx) e^x\) (since e^x is a root)
Step 3: Substitution & solve
Solve → \(A=1/2, B=-1\)Step 4: General solution
\[ y = C_1 e^x + C_2 e^{-x} + (1/2 x^2 - x) e^x \]Example 2: \( (D^2 + 4)y = x \cos 2x \)
Step 1: CF
\[ m^2 + 4 =0 \Rightarrow m = \pm 2i \quad y_c = C_1 \cos 2x + C_2 \sin 2x \]Step 2: PI
\[ y_p = (Ax + B)\cos 2x + (Cx + D)\sin 2x \]Step 3: Solve for coefficients
Solve → \(A=-1/8, B=0, C=1/8, D=0\)Step 4: General solution
\[ y = C_1 \cos 2x + C_2 \sin 2x + (-1/8 x \cos 2x + 1/8 x \sin 2x) \]Example 3: \( (D^2 - D - 2)y = x e^{-x} \)
Step 1: CF
\[ m^2 - m -2 =0 \Rightarrow m=2, -1 \quad y_c = C_1 e^{2x} + C_2 e^{-x} \]Step 2: PI
\[ y_p = (Ax^2 + Bx) e^{-x} \]Step 3: Solve
Solve → \(A=1/2, B=0\)Step 4: General solution
\[ y = C_1 e^{2x} + C_2 e^{-x} + (1/2 x^2 e^{-x}) \]Example 4: \( (D^2 + D)y = x \sin x \)
Step 1: CF
\[ m^2 + m =0 \Rightarrow m=0,-1 \quad y_c = C_1 + C_2 e^{-x} \]Step 2: PI
\[ y_p = (Ax + B) \sin x + (Cx + D) \cos x \]Step 3: Solve
Solve → \(A=-1, B=0, C=0, D=0\)Step 4: General solution
\[ y = C_1 + C_2 e^{-x} - x \sin x \]Example 5: \( (D^2 - 4)y = e^x \cos x \)
Step 1: CF
\[ m^2 -4 =0 \Rightarrow m = \pm 2 \quad y_c = C_1 e^{2x} + C_2 e^{-2x} \]Step 2: PI
\[ y_p = e^x (A \cos x + B \sin x) \]Step 3: Solve
Solve → \(A=1/5, B=-2/5\)Step 4: General solution
\[ y = C_1 e^{2x} + C_2 e^{-2x} + e^x (1/5 \cos x - 2/5 \sin x) \]Example 6: \( (D^2 - D)y = x e^x \)
Step 1: CF
\[ m^2 - m =0 \Rightarrow m = 0,1 \quad y_c = C_1 + C_2 e^x \]Step 2: PI (RHS = x e^x)
Trial: \(y_p = (Ax^2 + Bx) e^x\) (since e^x is a root)
Step 3: Derivatives & substitution
\[ y_p' = (2Ax + B)e^x + (Ax^2 + Bx)e^x = (Ax^2 + (2A+B)x + B)e^x \] \[ y_p'' = ... \text{(compute carefully)} \]Solve → \(A = 1, B=-2\)
Step 4: General solution
\[ y = C_1 + C_2 e^x + (x^2 - 2x)e^x \]Example 7: \( (D^2 + 1)y = x \cos x \)
Step 1: CF
\[ m^2 + 1=0 \Rightarrow m=\pm i \quad y_c = C_1 \cos x + C_2 \sin x \]Step 2: PI
\[ y_p = (Ax + B)\cos x + (Cx + D)\sin x \]Step 3: Substitution & solve
\[ A=0, B=0, C=-1/2, D=0 \]Step 4: General solution
\[ y = C_1 \cos x + C_2 \sin x - \frac{1}{2} x \sin x \]Example 8: \( (D^2 - 4)y = x e^{-2x} \)
Step 1: CF
\[ m^2 -4=0 \Rightarrow m = \pm 2 \quad y_c = C_1 e^{2x} + C_2 e^{-2x} \]Step 2: PI
\[ y_p = (Ax + B) e^{-2x} \]Step 3: Substitution
Solve → \(A = 1/2, B=0\)Step 4: General solution
\[ y = C_1 e^{2x} + C_2 e^{-2x} + \frac{1}{2} x e^{-2x} \]Example 9: \( (D^2 + 3D + 2)y = x e^{-x} \)
Step 1: CF
\[ m^2 + 3m +2 =0 \Rightarrow m=-1, -2 \quad y_c = C_1 e^{-x} + C_2 e^{-2x} \]Step 2: PI
\[ y_p = (Ax + B) e^{-x} \] (since e^{-x} is a root, multiply by x → \(y_p = Ax^2 e^{-x} + Bx e^{-x}\))Step 3: Solve for A,B
\[ A=1/2, B=0 \]Step 4: General solution
\[ y = C_1 e^{-x} + C_2 e^{-2x} + \frac{1}{2} x^2 e^{-x} \]Example 10: \( (D^2 - D -2)y = e^x \sin x \)
Step 1: CF
\[ m^2 - m -2 =0 \Rightarrow m = 2, -1 \quad y_c = C_1 e^{2x} + C_2 e^{-x} \]Step 2: PI
\[ y_p = e^x(A \sin x + B \cos x) \]Step 3: Substitution & solve
Solve → \(A = -2/5, B = 1/5\)Step 4: General solution
\[ y = C_1 e^{2x} + C_2 e^{-x} + e^x \left(-\frac{2}{5} \sin x + \frac{1}{5} \cos x\right) \]Method of Variation of Parameters
The method of variation of parameters is a powerful and general technique for solving inhomogeneous linear ordinary differential equations (ODEs). It's particularly useful when other methods, like the method of undetermined coefficients, do not apply.\\ The main conditions for applying the method of variation of parameters are:\\ The differential equation must be linear and non-homogeneous. The equation must be of the form \\ \[ y"+P(x)y'+Q(x)y=f(x)\] or a higher-order equivalent. The key is that the dependent variable and its derivatives appear only to the first power and are not multiplied by each other. The associated homogeneous equation must be solvable. You need to be able to find a fundamental set of linearly independent solutions \( y_1,y_2,y_3...) sponding homogeneous equation \[ y"+P(x)y'+Q(x)y=f(x)\]. This is a prerequisite for the method, as the particular solution is constructed from these homogeneous solutions. The coefficients and the non-homogeneous term, f(x), must be "reasonably continuous" functions. While the coefficients P(x) and Q(x) can be functions of x (not just constants), they need to be continuous over the interval of interest. This is a key advantage over the method of undetermined coefficients, which is limited to constant coefficients. The integrals involved must be solvable. The method requires you to compute integrals to find the unknown functions. If these integrals cannot be expressed in terms of elementary functions, the solution may have to be left in an integral form. In contrast, the method of undetermined coefficients is a faster, simpler alternative, but it only works when: The differential equation has constant coefficients. The non-homogeneous term, f(x), is of a specific, limited form (e.g., a polynomial, exponential, sine, cosine, or a combination of these). Variation of parameters is a general method that always works in theory as long as you can find the homogeneous solution and solve the resulting integrals, making it a reliable tool when undetermined coefficients is not an option.We solve non-homogeneous second-order linear differential equations of the form \[ y'' + P(x)y' + Q(x)y = R(x). \] If \(y_1, y_2\) are solutions of the homogeneous equation, then \[ u_1' = -\frac{R(x)y_2}{W}, \quad u_2' = \frac{R(x)y_1}{W}, \quad W = y_1y_2' - y_2y_1', \] and the particular solution is \[ y_p = u_1 y_1 + u_2 y_2. \]
Problem
Solve
\[
y'' - y = e^{2x}.
\]
1. Homogeneous solution: \(y_1 = e^x, \; y_2 = e^{-x}\).
2. Wronskian:
\[
W = e^x(-e^{-x}) - e^{-x}(e^x) = -2.
\]
3. Compute parameters:
\[
u_1' = \tfrac{e^x}{2}, \quad u_2' = -\tfrac{1}{2}e^{3x}.
\]
4. Integrate:
\[
u_1 = \tfrac{1}{2}e^x, \quad u_2 = -\tfrac{1}{6}e^{3x}.
\]
5. Particular solution:
\[
y_p = \tfrac{1}{3}e^{2x}.
\]
6. General solution:
\[
y = C_1 e^x + C_2 e^{-x} + \tfrac{1}{3} e^{2x}.
\]
Problem
Solve
\[
y'' + y = \sin x.
\]
1. Homogeneous solution: \(y_1 = \cos x, \; y_2 = \sin x\).
2. Wronskian:
\[
W = 1.
\]
3. Compute parameters:
\[
u_1' = -\sin^2 x, \quad u_2' = \sin x \cos x.
\]
4. Integrate:
\[
u_1 = -\tfrac{1}{2}x + \tfrac{1}{2}\sin x \cos x, \quad
u_2 = \tfrac{1}{2}\sin^2 x.
\]
5. Particular solution:
\[
y_p = -\tfrac{1}{2}x \cos x + \tfrac{1}{2}\sin x.
\]
6. General solution:
\[
y = C_1 \cos x + C_2 \sin x - \tfrac{1}{2}x \cos x + \tfrac{1}{2}\sin x.
\]