Answer to Question #122608 in Differential Equations for jessica

Question

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation and a particular solution yp(x) of the given nonhomogeneous equation.

y'' − 3y' + 2y = 7e^3x; y1 = e^x

y1(x) = ____
yp(x) = _____

Answer 1

$y''-3y'+2y=7e^{3x}, y_1=e^x.$

We use a substitution: $y=y_1z=e^xz$

$y'=(e^xz)'=e^x(z+z').$

$y''=[e^x(z+z')]'=e^x(z+2z'+z'').$

Substituting the derivatives in the equation and reducing by $e^x$ , we have

$z+2z'+z''-3(z+z')+2z=7e^{2x}\to z''-z'=7e^{2x}$

We use a substitution: $z'=u.$

$\frac{du}{dx}-u=7e^{2x}$. Let $t(x)=e^{-x}$ . Multiply both sides by $t(x)$.

$e^{-x}u'-\frac{d(e^{-x})}{dx}u=7e^x \to \frac{d(e^{-x})u}{dx}=7e^{x}$

$\int \frac{d(e^{-x}u)}{dx}dx=\int7e^{x}dx \to e^{-x}u=7e^{x}+C_1\to u(x)=e^x(7e^{x}+C_1)$

$z= \int udx= \int e^{x}(7e^{x}+C_1)=\frac{7e^{2x}}{2}+C_1e^x+C_2$

$y=e^xz=e^x(\frac{7e^{2x}}{2}+C_1e^x+C_2)$

Answer: $y_2=C_2e^{2x}$, the general solution is $y=e^x(\frac{7e^{2x}}{2}+C_1e^x+C_2)$.