Respuesta :
Answer:
[tex]\therefore y_2(x)=-\frac{e^{-6x}}{8}[/tex]
The general solution is
[tex]y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}[/tex]
Step-by-step explanation:
Given differential equation is
y''-4y'+4y=0
and [tex]y_1(x)=e^{2x}[/tex]
To find the [tex]y_2(x)[/tex] we are applying the following formula,
[tex]y_2(x)=y_1(x)\int \frac{e^{-\int P(x) dx}}{y_1^2(x)} \ dx[/tex]
The general form of equation is
y''+P(x)y'+Q(x)y=0
Comparing the general form of the differential equation to the given differential equation,
So, P(x)= - 4
[tex]\therefore y_2(x)=e^{2x}\int \frac{e^{-\int 4dx}}{(e^{2x})^2}dx[/tex]
[tex]=e^{2x}\int \frac{e^{-4x}}{e^{4x}}dx[/tex]
[tex]=e^{2x}\int e^{-4x-4x} \ dx[/tex]
[tex]=e^{2x}\int e^{-8x} \ dx[/tex]
[tex]=e^{2x}. \frac{e^{-8x}}{-8}[/tex]
[tex]=-\frac{e^{-6x}}{8}[/tex]
[tex]\therefore y_2(x)=-\frac{e^{-6x}}{8}[/tex]
The general solution is
[tex]y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}[/tex]
