What is the solution for y^{''}-4y^'=6t ?

The solution for y^{''}-4y^'=6t is y=c_{1}e^{4t}+c_{2}-(3t^2)/4-(3t)/8

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y^{''}-4y^'=6t

y^{^{''}} - 4 y^{^{'}} = 6 t

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

Solution steps

y^{''} (t) - 4 y^{'} (t) = 6 t

Solve linear ODE:

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

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

\frac{7 s ^{3} - 9 s + 1}{3 s ^{5}}

\int \frac{3}{x ( x ^{4} + 1 )} d x

\frac{\partial}{\partial z} (cos (x y))

x y y^{^{'}} - y^{2} = x^{4}

e^{x} \cdot ln (x)

What is the solution for y^{''}-4y^'=6t ?
The solution for y^{''}-4y^'=6t is y=c_{1}e^{4t}+c_{2}-(3t^2)/4-(3t)/8