Question

Find the general solution of the following nonhomogeneous second order differential equation: y" - 4y = e^2x

Answer 1

Answer:

Solution is $y=c_{1}e^{2x}+c_{2}e^{-2x}+\frac{1}{4}xe^{2x}$

Step-by-step explanation:

the given equation y''-4y$=e^{2x}$ can be written as

$D^{2}y-4y=e^{2x}\\\\(D^{2}-4)y=e^{2x}$

The Complementary function thus becomes

y=c_{1}e^{m_{1}x}+c_{2}e^{m_{2}x}

where $m_{1} , m_{2}$ are the roots of the $D^{2}-4$

The roots of $D^{2}-4$ are +2,-2 Thus the comlementary function becomes

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

here $c_{1},c_{2}$ are arbitary constants

Now the Particular Integral becomes using standard formula

$y=\frac{e^{ax}}{f(D)}\\\\y=\frac{e^{ax}}{f(a)} (f(a)\neq 0)\\\\y=x\frac{e^{ax}}{f'(a)}(f(a)=0)$

$y=\frac{e^{2x}}{D^{2}-4}\\\\y=\frac{e^{2x}}{(D+2)(D-2)}\\\\y=\frac{1}{D-2}\times \frac{e^{2x}}{2+2}\\\\y=\frac{1}{4}\times \frac{e^{2x}}{D-2}\\\\y=\frac{1}{4}xe^{2x}$

Hence the solution is = Complementary function + Particular Integral

Thus Solution becomes $y=c_{1}e^{2x}+c_{2}e^{-2x}+\frac{1}{4}xe^{2x}$