Question

Solve the partial differential equation (3D² +10DD'+ 3D ^1^2)z^ =e^x-y

Answer 1

Since, the given equation is unclear, we assume it as:

$\quad Solve \left(D^{2}+4 D+3\right) y=e^{-3 x}$

Solution: Given equation is

$\left(D^{2}+4 D+3\right) y=e^{-3 x}$

A E is $\quad D^{2}+4 D+3=0$

$\begin{aligned} &(D+1)(D+3)=0 \quad D=-1, \quad D=-3 \\ \therefore \quad & C F=c_{1} e^{-x}+c_{2} e^{-3 x} \\ & P I=\frac{1}{(D+1)(D+3)} e^{-3 x} \end{aligned}$

put D=-3 in f(D) then

$\begin{aligned} P I &=\frac{1}{0} e^{-3 x} \quad \therefore \text { method fails } \\ \therefore \quad P I &=\frac{e^{-3 x}}{(-3+1)(D-3+3)}\{1\}=\frac{e^{-3 x}}{(-2)} \frac{1}{D}\{1\}=\frac{e^{-3 x}}{-2} x \\ G . S . &=C F+P I \end{aligned}$

$y=c_{1} e^{-x}+c_{2} e^{-3 x}-\frac{e^{-3 x}}{2} x$