Question

Solve the initial value problem

y"-5y'+6y=2e^x, y(0)=1, y'(0)=1.

Answer 1

Related homogeneous (complementary) equation

$$y''-5y'+6y=0$$

Characteristic (auxiliary) equation

$$r^2-5r+6=0$$

$$(r-2)(r-3)=0$$

$$r_1=2, r_2=3$$

The general solution of the homogeneous equation is

$$y_h=c_1e^{2x}+c_2e^{3x}$$

Find the particular solution of the nonhomogeneous differential equation

$$y_p=Ae^x$$

Then

$$y_p'=Ae^x$$

$$y_p''=Ae^x$$

Substitute

$$Ae^x-5Ae^x+6Ae^x=2e^x$$

$$A=1$$

$$y_p=e^x$$

The general solution of the nonhomogeneous equation is

$$y=y_h+y_p$$

$$y=c_1e^{2x}+c_2e^{3x}+e^x$$

Initial conditions

$y(0)=1:$

$$1=c_1e^{2(0)}+c_2e^{3(0)}+e^0=>c_1+c_2=0$$

$y'(0)=1$

$$y'=2c_1e^{2x}+3c_2e^{3x}+e^x$$

$$1=2c_1e^{2(0)}+3c_2e^{3(0)}+e^0=>2c_1+3c_2=0$$

$$c_2=-c_1$$

$$2c_1-3c_1=0$$

$$c_1=0, c_2=0$$

The solution of the given initial value problem is

$$y=e^x$$