Answer
$y=c_1.e^{x}+c_2.x.e^{x}+e^{2.x}$
Work Step by Step
We need to find the general solution of:
$y''-2.y’+y=e^{2.x}$
First we solve the homogeneous equation associated to the general one:
$y''-2.y'+y=0$
The caracteristic equation is:
$m^2-2.m+1=0$
Solving for m:
$Δ=(-2)^2-4.1.1=0$
$m_1=m_2=\frac{-b}{2.a}=\frac{2}{2}=1$
By theorem of homogeneous linear equations, we have:
$y_h=c_1.e^{x}+c_2.x.e^{x}$
Now we’ll find the particular solution to
$y''-2.y’+y=e^{2.x}$
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 sin informations inside (*), we have:
$y''-2.y’+y=4.a.e^{2.x}-2.2.a.e^{2.x}+a.e^{2.x}=e^{2.x}$
Then:
$a.e^{2.x}=e^{2.x}$
By equality:
$a=1$
Then:
$y_p=e^{2.x}$
Therefore:
$y=y_h+y_p=c_1.e^{x}+c_2.x.e^{x}+e^{2.x}$
Variation of parameters)
We have
$y_1=e^{x}$
$y_2=x.e^{x}$
The Wronskian is:
$W=\begin{vmatrix}
e^{x} & x.e^{x} \\
e^{x} & e^{x}+x.e^{x}
\end{vmatrix}=e^{2.x}+x.e^{2.x}-x.e^{2.x}=e^{2.x}$
Then, we need to find the solution function as an integral, in order to sum with $y_h$. We have:
$\psi(x)=-y_1\int{\frac{y_2.e^{2.x}}{e^{2.x}}}dx+y_2\int{\frac{y_1.e^{2.x}}{e^{2.x}}}dx$
Substituing the function, we have:
$\psi(x)=-e^{x}\int{\frac{x.e^{x}.e^{2.x}}{e^{2.x}}}dx+x.e^{x}\int{\frac{e^{x}.e^{2.x}}{e^{2.x}}}dx$
By integration by parts in first integral:
$\psi(x)=-e^{x}\int{x.e^{x}}dx+x.e^{x}\int{e^{x}}dx=-e^{x}.(x.e^{x}-\int{e^{x}}dx)+x.e^{2x}=-x.e^{2x}+e^{2.x}+x.e^{2.x}=e^{2.x}$
Therefore, the solution is:
$y=y_h+\psi(x)=c_1.e^{x}+c_2.x.e^{x}+e^{2.x}$
