Question

[D] y''-25y=0; y1=e^5x the indicated function y1(x) is a solution of the given differential eqution.Use reduction of order or formula as instructed, to find a second solution y2(x).

Answer 1

$$y''+(0)y'+(-25)y=0$$

$$P(x)=0, Q(x)=-25$$

$$y_1(x)=e^{5x}$$

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

$$y_2(x)=e^{5x}\int \dfrac{e^{-\int (0)dx}}{(e^{5x})^2}dx$$

$$=Ce^{5x}\int e^{-10x}dx=-\dfrac{C}{10}e^{5x}e^{-10x}$$

$$=C_1e^{-5x}$$

$y_2(x)=e^{-5x}$ is a solution of the given differential eqution.