Answer
$y=c_1.e^{2.x}.sin(x)+c_2.e^{2.x}.cos(x)+e^{2.x}$
Work Step by Step
The equation can be written as:
$y''-4.y’+5y=e^{2.x}$
First we solve the homogeneous equation associated to the general one:
$y''-4.y'+5y=0$
The caracteristic equation is:
$m^2-4.m+5=0$
Solving for m:
$Δ=(-4)^2-4.1.5=-4$
$m_1=\frac{4+2.i}{2.1}=2+i$
$m_2=\frac{4-2.i}{2.1}=2-i$
By theorem of homogeneous linear equations, we have:
$y_h=c_1.e^{2.x}.sin(x)+c_2.e^{2.x}.cos(x)$
Now we’ll find the particular solution to
$y''-4.y’+5y=e^{2.x}$
By undetermined coefficients, taking the general possible solution. We’ll take:
$y=a.e^{2.x}$
Differentiating, we have:
$y'=2.a.e^{2.x}$
$y''=4.a.e^{2.x}$
Now, putting the informations inside (*), we have:
$y''-4.y’+5y=4.a.e^{2.x}-4.2.a.e^{2.x}+5.a.e^{2.x}=e^{2.x}$
Then:
$a.e^{2.x}=e^{2.x}$
By equality:
$a=1$
Then:
$y_p=e^{2.x}$
Therefore:
$y=y_h+y_p=c_1.e^{2.x}.sin(x)+c_2.e^{2.x}.cos(x)+e^{2.x}$