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snow_tiger [21]
3 years ago
15

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
Mathematics
1 answer:
lisabon 2012 [21]3 years ago
4 0

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}

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