Question

D^2y/dx^2 - 2 dy/dx + 4y = e^x*cosx

Answer 1

First, the characteristic solution to the homogeneous part:

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-2\dfrac{\mathrm dy}{\mathrm dx}+4y=0$

has characteristic equation

$r^2-2r+4=0$

with roots $r=1\pm\sqrt3i$. So the characteristic solution has the form

$y_c=e^x(C_1\cos\sqrt3x+C_2\sin\sqrt3x)$

Now for the nonhomogeneous part. Using the method of undetermined coefficients, we can try a particular solution of the form

$y_p=e^x(a\cos x+b\sin x)$

which has derivatives

$\dfrac{\mathrm dy_p}{\mathrm dx}=e^x((a+b)\cos x+(-a+b)\sin x)$
$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=2e^x(b\cos x-a\sin x)$

Substituting $y_p$ and its derivatives into the equation gives

$2e^x(b\cos x-a\sin x)-2e^x((a+b)\cos x+(-a+b)\sin x)+4e^x(a\cos x+b\sin x)=e^x\cos x$
$2e^x(a\cos x+b\sin x)=e^x\cos x$

which means

$\begin{cases}2a=1\\2b=0\end{cases}\implies a=\dfrac12,b=0$

so that the particular solution must be

$y_p=\dfrac12e^x\cos x$

Now the general solution will be

$y=y_c+y_p$
$y=e^x(C_1\cos\sqrt3x+C_2\sin\sqrt3x)+\dfrac12e^x\cos x$