find the particular solution of y'' + y' = x + e^x satisfying y(0)=0 and y'(0)=1

Question

myininaya · Answer

$$y=Ax^2+Bx+Cxe^x+De^x$$
$$y'=2Ax+B+Ce^x+Cxe^x+De^x$$
$$y''=2A+Ce^x+Ce^x+Cxe^x+De^x$$
so we have
y''+y'=
$$[xe^x(C)+e^x(2C+D)+2A]+[xe^x(C)+e^x(C+D)+x(2A)+B]$$
$$=xe^x(2C)+e^{x}(3C+2D)+x(2A)+(2A+B)$$
this is suppose to equal x+e^x
so we have
$$2C=0; 3C+2D=1; 2A=1; 2A+B=0$$

$$C=0;D=\frac{1}{2}; A=\frac{1}{2}; B=-1$$

myininaya · Answer

$$y=\frac{1}{2}x^2-x+\frac{1}{2}e^x$$

myininaya · Answer

+constant

myininaya · Answer

you also need to find the homogeneous solution

myininaya · Answer

$$y''+y'=0$$
solve this

myininaya · Answer

$$r^2+r=0 => r(r+1)=0 => r=0; r=-1=>y_h=c_1+c_2e^{-t}$$

myininaya · Answer

$$y=\frac{1}{2}x^2-x+\frac{1}{2}e^x+c_2e^{-t}+c$$

myininaya · Answer

oops t is x

myininaya · Answer

now you need to apply your initial conditions