Question

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

Answer 1

\[y=Ax^2+Bx+Cxe^x+De^x\]
\[y'=2Ax+B+Ce^x+Cxe^x+De^x\]
\[y''=2A+Ce^x+Ce^x+Cxe^x+De^x\]
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
\[2C=0; 3C+2D=1; 2A=1; 2A+B=0\]

\[C=0;D=\frac{1}{2}; A=\frac{1}{2}; B=-1\]

Answer 2

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

Answer 3

+constant

Answer 4

you also need to find the homogeneous solution

Answer 5

\[y''+y'=0\]
solve this

Answer 6

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

Answer 7

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

Answer 8

oops t is x

Answer 9

now you need to apply your initial conditions