Particular solution of linear higher order ODE
Finding the $y_p$ for for the particular component of the general solution
Particular Integral P.I. #
$y_p(x)$ is also called as particular integral/ particular solution of the non-homogeneous equation.
$$ \begin{align*} P.I. of\\ f(D)y=g(x):\\ \\ y_{p}=\frac{1}{f(D)}g(x) \end{align*} $$
$\frac{1}{f(D)}$ is the inverse operator of $f(D)$
Particular Integral of Functions #
case 1 #
$$ \begin{align*} y_p=\frac{1}{D-\alpha}g(x)=e^{\alpha x}\int e^{-\alpha x}g(x)dx\\ \\
(D-\alpha)y=g(x)\\ \dd{y}{x}-\alpha y=g(x); I.F.=e^{-\alpha x}\\ e^{-\alpha x}y=\int e^{-\alpha x}g(x)dx\\ \\ [\because [1]]\\ \\ Hence\\ proved \end{align*} $$ 1
case 2: Exponential shift #
$$ \begin{align*} \frac{1}{f(D)}e^{\alpha x}V(x)=e^{\alpha x} \frac{1}{f(D+\alpha)} V(x)\\ \\ \\ Can\\ be\\ proved\\ with: D^{r}(e^{ax}u)=e^{ax}(D+\alpha)^{r}u \end{align*} $$
case 1 can also be proved using case 2 #
$$ \begin{align*} \frac{1}{D-\alpha}g(x)=\frac{1}{D-\alpha}e^{\alpha x}e^{-\alpha x}g(x)\\ =e^{\alpha x} \frac{1}{D-\alpha+\alpha}e^{-\alpha x}g(x)\\ =e^{\alpha x} \frac{1}{D}e^{-\alpha x}g(x) \end{align*} $$
case 3: g(x)=$e^{ax}$ #
$$ \frac{1}{f(D)}e^{ax}=\frac{1}{f(a)}e^{ax} $$
If f(a)≠0. If f(a)=0,
$$ \begin{align*} \frac{1}{f(D)}e^{ax}=\frac{1}{(D-a)^{p}\Phi(D)}e^{ax}\\ \frac{1}{\Phi (a)} \frac{1}{(D-a)^{p}}e^{ax}\\ \frac{1}{\Phi (a)}e^{ax} \frac{1}{(D-a+a)^{p}}1\\ \frac{1}{\Phi (a)}e^{ax} \frac{x^{p}}{p!} \end{align*} $$
Alternate solution:
$\frac{1}{f(D)}e^{ax}=x\frac{1}{f’(D)}e^{ax}$ or if f’(a)=0, $x^{2} \frac{1}{f’’(D)}e^{ax}$
new case g(x)=sin(ax) or cos(ax) #
If f(D) is even function. i.e. can be written as F(D^2) $$ \begin{align*} \frac{1}{F(D^{2})}\sin(ax)\\ \\ \\ P.I. &= \frac{1}{F(-a^{2})}(\sin ax) \end{align*} $$
If f(D) contains odd powers of D f(D)=f_1(D^2)+f_2(D)
$$ \begin{align*} \frac{1}{f_{1}(D^{2})+f_{2}(D)} \sin(ax)\\ \frac{1}{f_{1}(-a^{2})+f_{2}(D)} \sin(ax)\
\end{align*} $$ Examples We simplify and multiply and divide by f_3(D), where it contains odd powers of D to make the denominator even. $$ \begin{align*} \frac{f_3(D)}{(f_{1}(-a^{2})+f_{2}(D))f_{3}(D)} \sin(ax)\\\ \frac{f_3(D)}{\Phi (-a^{2})} \sin(ax) \end{align*} $$ Examples
If denominator is 0 Examples
New case g(x)=x^m #
$$ \begin{align*} P.I. = \frac{1}{f(D)} x^m=[f(D)]^{-1}x^m \end{align*}
$$
Expand f(D)⁻¹ is ascending powers of D as per the terms in $D^m$ and operate on x^m term by term. Since the (m+1)th and higher derivatives of $x^m$ are 0, we need not consider terms beyond $D^m$
Basically have D^0, D^1,…D^m terms.
Expansions #
$$ \begin{align*} \frac{1}{1-x}=1+x+x^{2}+x^{3}+x^{4}\\ \\ \frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4} \end{align*} $$
g(x)=xV(x) #
$$ \text{P.I.= } \frac{1}{f(D)}xV=\left(x-\frac{f’(D)}{f(D)}\right) \frac{1}{f(D)} V $$