Answer
$y=-sin(t)+cos(t)+t^3-1$
Work Step by Step
Step 1. Given $y^{(4)}=-sin(t)+cos(t)$, we can find its general antiderivatives as $y'''=cos(t)+sin(t)+C$ where $C$ is a constant.
Step 2. Use the initial condition $y'''(0)=7$, we have $cos(0)+sin(0)+C=7$ which gives $C=6$ and we get $y'''=cos(t)+sin(t)+6$
Step 3. Find the general antiderivatives of the above to get $y''=sin(t)-cos(t)+6t+D$ where $D$ is a constant.
Step 4. Using the initial condition $y''(0)=-1$, we have $sin(0)-cos(0)+6(0)+D=-1$ which gives $D=0$ and we get $y''=sin(t)-cos(t)+6t$
Step 5. Find the general antiderivatives of the above to get $y'=-cos(t)-sin(t)+3t^2+E$, where $E$ is a constant.
Step 6. Using the initial condition $y'(0)=-1$, we have $-cos(0)-sin(0)+3(0)^2+E=-1$ which gives $E=0$ and we get $y'=-cos(t)-sin(t)+3t^2$
Step 7. Finally, find the general antiderivatives of the above to get $y=-sin(t)+cos(t)+t^3+F$ where $F$ is a constant.
Step 8. Using the initial condition $y(0)=0$, we have $-sin(0)+cos(0)+(0)^3+F=0$, which gives $F=-1$ and we get $y=-sin(t)+cos(t)+t^3-1$ as our answer.
