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

`y_2=xe^(2x)`

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 (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 (e^(-\int p(x)\, dx))/(y^2_1(x))\, dx\\\\y_2=e^(2x)\int (e^(4\int 1\, dx))/((e^(2x))^2)\, dx\\\\y_2=e^(2x)\int (e^(4x))/(e^(4x))\, dx\\\\y_2=e^(2x)\int 1\, dx\\\\y_2=xe^(2x)`

So, we get that

`y_2=xe^(2x)`