What is the solution for 8t(dy)/(dt)+y=t^4 ?

The solution for 8t(dy)/(dt)+y=t^4 is y=(t^4)/(33)+(c_{1})/(\sqrt[8]{t)}

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8t(dy)/(dt)+y=t^4

8 t \frac{d y}{d t} + y = t^{4}

y = \frac{t ^{4}}{33} + \frac{c _{1}}{8 t}

Solution steps

8 t \frac{d y}{d t} + y = t^{4}

Solve linear ODE:

y = \frac{t ^{4}}{33} + \frac{c _{1}}{8 t}

y = \frac{t ^{4}}{33} + \frac{c _{1}}{8 t}

\frac{\partial}{\partial y} (x + x^{2} + y^{2})

\frac{d}{d x} (\frac{6 x - 1}{4 x + 3})

\frac{\partial}{\partial v} (u v^{2})

\frac{d}{d x} (\frac{1}{( x + 8 ) ^{2}})

\int \frac{1}{x ^{2} ( x - 2 )} d x

What is the solution for 8t(dy)/(dt)+y=t^4 ?
The solution for 8t(dy)/(dt)+y=t^4 is y=(t^4)/(33)+(c_{1})/(\sqrt[8]{t)}