Answer
$y=-cos(x)+\frac{1}{2}sin(2x)+\frac{2}{3}x^3+4$
Work Step by Step
Step 1. Given $y^{(4)}=-cos(x)+8sin(2x)$, we can find its general antiderivatives as $y'''=-sin(x)-4cos(2x)+C$ where $C$ is a constant.
Step 2. Using the initial condition $y'''(0)=0$, we have $-sin(0)-4cos(0)+C=0$, which gives $C=4$ and we get $y'''=-sin(x)-4cos(2x)+4$
Step 3. Find the general antiderivatives of the above to get $y''=cos(x)-2sin(2x)+4x+D$ where $D$ is a constant.
Step 4. Using the initial condition $y''(0)=1$, we have $cos(0)-2sin(0)+4(0)+D=1$, which gives $D=0$ and we get $y''=cos(x)-2sin(2x)+4x$
Step 5. Find the general antiderivatives of the above to get $y'=sin(x)+cos(2x)+2x^2+E$ where $E$ is a constant.
Step 6. Using the initial condition $y'(0)=1$, we have $sin(0)+cos(0)+2(0)^2+E=1$, which gives $E=0$ and we get $y'=sin(x)+cos(2x)+2x^2$
Step 7. Finally, find the general antiderivatives of the above to get $y=-cos(x)+\frac{1}{2}sin(2x)+\frac{2}{3}x^3+F$ where $F$ is a constant.
Step 8. Using the initial condition $y(0)=3$, we have $-cos(0)+\frac{1}{2}sin(0)+\frac{2}{3}(0)^3+F=3$ which gives $F=4$ and we get $y=-cos(x)+\frac{1}{2}sin(2x)+\frac{2}{3}x^3+4$ as our answer.