Answer to Question #207413 in Differential Equations for Abigail

Question

how to solve y''-y'-12y=e^4x

Answer 1

$$y''-y'-12y=e^{4x}$$

Write the related homogeneous or complementary equation:

$$y''-y'-12y=0$$

The general solution of a nonhomogeneous equation is the sum of the general solution $y_h(x)$ of the related homogeneous equation and a particular solution $y_p(x)$ of the nonhomogeneous equation:

$$y(x)=y_h(x)+y_p(x)$$

Consider a homogeneous equation 

$$y''-y'-12y=0$$

Write the characteristic (auxiliary) equation:

$$r^2-r-12=0$$

$$(r-4)(r+3)=0$$

$$r_1=4, r_2=-3$$

The general solution of the homogeneous equation is

$$y_h(x)=C_1e^{4x}+C_2e^{-3x}$$

Method of undetermined coefficients

Let the general solution of a second order homogeneous differential equation be

$$y(x)=C_1(x)e^{4x}+C_2(x)e^{-3x}$$

The unknown functions $C_1(x)$ and $C_2(x)$ can be determined from the system of two equations:

$$C_1'e^{4x}+C_2'e^{-3x}=0$$

$$C_1'(4e^{4x})+C_2'(-3e^{-3x})=e^{4x}$$

$$C_2'e^{-3x}=-C_1'e^{4x}$$

$$4C_1'e^{4x}+3C_1'e^{4x}=e^{4x}$$

$$C_1'=\dfrac{1}{7}$$

$$C_2'=-\dfrac{1}{7}e^{7x}$$

$$C_1=\dfrac{1}{7}x+c_3$$

$$C_2=-\dfrac{1}{49}e^{7x}+c_2$$

The general solution of a second order homogeneous differential equation be

$$y(x)=c_1e^{4x}+c_2e^{-3x}+\dfrac{1}{7}xe^{4x}$$