What is the solution for h^{''}=40+4t ?

The solution for h^{''}=40+4t is h(t)=20t^2+(2t^3)/3+c_{1}t+c_{2}

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h^{''}=40+4t

h^{^{''}} = 40 + 4 t

h (t) = 20 t^{2} + \frac{2 t ^{3}}{3} + c_{1} t + c_{2}

Solution steps

h^{''} (t) = 40 + 4 t

Solve

h^{''} (t) = 40 + 4 t : h^{'} (t) = 40 t + 2 t^{2} + c_{1}

Solve

h^{'} (t) = 40 t + 2 t^{2} + c_{1} : h (t) = 20 t^{2} + \frac{2 t ^{3}}{3} + c_{1} t + c_{2}

h (t) = 20 t^{2} + \frac{2 t ^{3}}{3} + c_{1} t + c_{2}

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What is the solution for h^{''}=40+4t ?
The solution for h^{''}=40+4t is h(t)=20t^2+(2t^3)/3+c_{1}t+c_{2}