Question 2 Identify the solution to the initial value problem \[ \left\{\begin{array}{l} y^{\prime \prime \prime}(t)=24 t \\ y(0)=0 \\ y^{\prime}(0)=1 \\ y^{\prime \prime}(0)=2 \end{array}\right. \] $t^{4}+t^{2}+t$ $t^{4}+\frac{1}{2} t^{2}+t$ $24 t^{4}+2 t^{2}+t$ $t+24 t^{2}$

Step 1 ：Integrate \(y'''(t) = 24t\) with respect to t to get \(y''(t) = \int y'''(t) dt = \int 24t dt = 12t^2 + C1\)

Step 2 ：Apply the initial condition \(y''(0) = 2\) to find \(C1\): \(2 = 12*0^2 + C1\) which gives \(C1 = 2\). So, \(y''(t) = 12t^2 + 2\)

Step 3 ：Integrate \(y''(t) = 12t^2 + 2\) with respect to t to get \(y'(t) = \int y''(t) dt = \int (12t^2 + 2) dt = 4t^3 + 2t + C2\)

Step 4 ：Apply the initial condition \(y'(0) = 1\) to find \(C2\): \(1 = 4*0^3 + 2*0 + C2\) which gives \(C2 = 1\). So, \(y'(t) = 4t^3 + 2t + 1\)

Step 5 ：Integrate \(y'(t) = 4t^3 + 2t + 1\) with respect to t to get \(y(t) = \int y'(t) dt = \int (4t^3 + 2t + 1) dt = t^4 + t^2 + t + C3\)

Step 6 ：Apply the initial condition \(y(0) = 0\) to find \(C3\): \(0 = 0^4 + 0^2 + 0 + C3\) which gives \(C3 = 0\)

Step 7 ：So, the solution to the initial value problem is \(\boxed{y(t) = t^4 + t^2 + t}\)