Question

d²y/dx² - dy/dx= e^-x , y(0) =0, y'(0) =1'

Answer 1

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

Write the related homogeneous or complementary equation:

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

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 

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

Write the characteristic (auxiliary) equation:

$$r^2-r=0$$

$$r(r-1)=0$$

$$r_1=0, r_2=1$$

The general solution of the homogeneous equation is

$$y_h(x)=C_1+C_2e^{x}$$

Find the particular solution of the nonhomogeneous differential equation

$$y_p(x)=Ae^{-x}$$

$$y_p'=-Ae^{-x}$$

$$y_p''=Ae^{-x}$$

Substitute

$$Ae^{-x}-(-Ae^{-x})=e^{-x}$$

$$A=\dfrac{1}{2}$$

$$y_p(x)=\dfrac{1}{2}e^{-x}$$

The general solution of a second order homogeneous differential equation is

$$y(x)=C_1+C_2e^{x}+\dfrac{1}{2}e^{-x}$$

Initial conditions: $y(0)=0, y'(0)=1$

$$y(0)=C_1+C_2e^{0}+\dfrac{1}{2}e^{-0}=0$$

$$y'(x)=C_2e^{x}-\dfrac{1}{2}e^{-x}$$

$$y'(0)=C_2e^{0}-\dfrac{1}{2}e^{-0}=1$$

$$C_2=\dfrac{3}{2}$$

$$C_1=-2$$

The solution of the given differential equation is

$$y(x)=-2+\dfrac{3}{2}e^{x}+\dfrac{1}{2}e^{-x}$$