Answer
$y=\frac{3}{2}cosx+\frac{11}{2}sinx+\frac{1}{2}e^{x}+x^{3}-6x$
Work Step by Step
$y''+y=0$
Use auxiliary equation
$r^{2}+1=0$
$r^{2}=-1$
$r=±i$
$α=0$
$β=1$
$y_{c}=e^{αx}(c_{1}cosβx+c_{2}sinβx)$
$y_{c}=c_{1}cosx+c_{2}sinx$
The particular solution is of the form $y_{p}=Ae^{x}+Bx^{3}+Cx^{2}+Dx+E$
$y_{p}'=Ae^{x}+3Bx^{2}+2Cx+D$
$y_{p}''=Ae^{x}+6Bx+2C$
Plug back into the original equation
$(Ae^{x}+6Bx+2C)+(Ae^{x}+Bx^{3}+Cx^{2}+Dx+E)=e^{x}+x^{3}$
$(2A)e^{x}+x^{3}(B)+x^{2}(C)+x(6B+D)+(2C+E)=e^{x}+x^{3}$
$2A=1$
$A=\frac{1}{2}$
$B=1$
$C=0$
$6B+D=0$
$6(1)+D=0$
$D=-6$
$2C+E=0$
$2(0)+E=0$
$E=0$
Therefore, the particular solution is
$y_{p}=\frac{1}{2}e^{x}+x^{3}-6x$
$y=y_{c}+y_{p}$
$y=c_{1}cosx+c_{2}sinx+\frac{1}{2}e^{x}+x^{3}-6x$
First condition $y(0)=2$
$2=c_{1}cos(0)+c_{2}sin(0)+\frac{1}{2}e^{0}+(0)^{3}-6(0)$
$2=c_{1}+\frac{1}{2}$
$c_{1}=\frac{3}{2}$
Second condition $y'(0)=0$
$y'=-c_{1}sinx+c_{2}cosx+\frac{1}{2}e^{x}+3x^{2}-6$
$0=-c_{1}sin(0)+c_{2}cos(0)+\frac{1}{2}e^{0}+3(0)^{2}-6$
$0=c_{2}+\frac{1}{2}-6$
$c_{2}=\frac{11}{2}$
$y=\frac{3}{2}cosx+\frac{11}{2}sinx+\frac{1}{2}e^{x}+x^{3}-6x$
