Question

The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2, y2 = y1(x) e−∫P(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). y'' − 4y' + 4y = 0; y1 = e2x

Answer 1

Answer:

$\therefore y_2(x)=-\frac{e^{-6x}}{8}$

The general solution is

$y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}$

Step-by-step explanation:

Given differential equation is

y''-4y'+4y=0

and $y_1(x)=e^{2x}$

To find the $y_2(x)$ we are applying the following formula,

$y_2(x)=y_1(x)\int \frac{e^{-\int P(x) dx}}{y_1^2(x)} \ dx$

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

$\therefore y_2(x)=e^{2x}\int \frac{e^{-\int 4dx}}{(e^{2x})^2}dx$

           $=e^{2x}\int \frac{e^{-4x}}{e^{4x}}dx$

           $=e^{2x}\int e^{-4x-4x} \ dx$

            $=e^{2x}\int e^{-8x} \ dx$

            $=e^{2x}. \frac{e^{-8x}}{-8}$

           $=-\frac{e^{-6x}}{8}$

$\therefore y_2(x)=-\frac{e^{-6x}}{8}$

The general solution is

$y=c_1e^{2x}-c_2.\frac{e^{-6x}}{8}$