Answer
$y(x)=c_1\cos 2x+c_2\sin 2x+\frac{7}{5}e^x$
Work Step by Step
Given: $y''+4y=7e^x$
Solve the auxiliary equation for the differential equation.
$r^2+4=0$
Factor and solve for the roots.
$r_1=2i\\
r_2=-2i$
Roots are: $r_1=2i$, as a multiplicity of 1 and $r_2=-2i$, as a multiplicity of 1
The general solution is
$y(x)=c_1\cos 2x+c_2\sin 2x$
We have $F(x)=7e^x$
Obtain:
$(D-1)(D^2+4)^3y_p(x)=7e^x$
Therefore, the general solution for $(D-1)(D^2+4)^3y_p(x)=0$ is:
$y(x)=c_1\cos 2x+c_2\sin 2x+A_0e^x$
The trial solution can be computed as by plugging back into the given differential equation.
$y_p(x)=A_0e^x$
So, we have:
$(D-1)(D^2+4)^3y_p(x)=7e^x\\
(D-1)(D^2+4)^3(A_0e^x)=7e^x$
On comparing coefficients, we get:
$A_0=\frac{7}{5}$
Therefore, the general solution for the given differential equation is:
$y(x)=c_1\cos 2x+c_2\sin 2x+\frac{7}{5}e^x$
