What is the solution for y^{''}=4t^3+t ?

The solution for y^{''}=4t^3+t is y=(t^5)/5+(t^3)/6+c_{1}t+c_{2}

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y^{''}=4t^3+t

y^{^{''}} = 4 t^{3} + t

y = \frac{t ^{5}}{5} + \frac{t ^{3}}{6} + c_{1} t + c_{2}

Solution steps

y^{''} (t) = 4 t^{3} + t

Solve

y^{''} (t) = 4 t^{3} + t : y^{'} (t) = t^{4} + \frac{t ^{2}}{2} + c_{1}

Solve

y^{'} (t) = t^{4} + \frac{t ^{2}}{2} + c_{1} : y = \frac{t ^{5}}{5} + \frac{t ^{3}}{6} + c_{1} t + c_{2}

y = \frac{t ^{5}}{5} + \frac{t ^{3}}{6} + c_{1} t + c_{2}

What is the solution for y^{''}=4t^3+t ?
The solution for y^{''}=4t^3+t is y=(t^5)/5+(t^3)/6+c_{1}t+c_{2}