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

The solution for y^{''}-4y^'=3t is y=c_{1}e^{4t}+c_{2}-(3t^2)/8-(3t)/(16)

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

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

y = c_{1} e^{4 t} + c_{2} - \frac{3 t ^{2}}{8} - \frac{3 t}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} - \frac{3 t ^{2}}{8} - \frac{3 t}{16}

y = c_{1} e^{4 t} + c_{2} - \frac{3 t ^{2}}{8} - \frac{3 t}{16}

What is the solution for y^{''}-4y^'=3t ?
The solution for y^{''}-4y^'=3t is y=c_{1}e^{4t}+c_{2}-(3t^2)/8-(3t)/(16)