Question

The indicated function y1(x) is a solution of the given equation. Use reduction of order or this formula, y2 = y1(x) e−∫P(x) dx y12(x) dx , as instructed, to find a second solution y2(x). y'' − 4y' + 4y = 0; y1 = e2x y2 =

Answer 1

Answer:

$y_2=xe^{2x}$

Step-by-step explanation:

We have the differential equation  y'' − 4y' + 4y = 0,

so we get that is p(x)=-4.

We use the formula:

$y_2=y_1(x)\int \frac{e^{-\int p(x)\, dx}}{y^2_1(x)}\, dx$

We have that:

$y_1=e^{2x}$

we calculate:

$y_2=y_1(x)\int \frac{e^{-\int p(x)\, dx}}{y^2_1(x)}\, dx\\\\y_2=e^{2x}\int \frac{e^{4\int 1\, dx}}{(e^{2x})^2}\, dx\\\\y_2=e^{2x}\int \frac{e^{4x}}{e^{4x}}\, dx\\\\y_2=e^{2x}\int 1\, dx\\\\y_2=xe^{2x}$

So, we get that

$y_2=xe^{2x}$