What is the solution for 6t(dy)/(dt)+y=t^3 ?

The solution for 6t(dy)/(dt)+y=t^3 is y=(t^3)/(19)+(c_{1})/(\sqrt[6]{t)}

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6t(dy)/(dt)+y=t^3

6 t \frac{d y}{d t} + y = t^{3}

y = \frac{t ^{3}}{19} + \frac{c _{1}}{6 t}

Solution steps

6 t \frac{d y}{d t} + y = t^{3}

Solve linear ODE:

y = \frac{t ^{3}}{19} + \frac{c _{1}}{6 t}

y = \frac{t ^{3}}{19} + \frac{c _{1}}{6 t}

x^{2} + 4 y^{2}

y^{^{'}} = \frac{( 3 t ^{2} - e ^{t} )}{2 y - 5}

x \to 5 lim (2 e^{x} - 2 ln (x))

\frac{d}{d t} (- 2 t)

\int tan^{2} (1) d x

What is the solution for 6t(dy)/(dt)+y=t^3 ?
The solution for 6t(dy)/(dt)+y=t^3 is y=(t^3)/(19)+(c_{1})/(\sqrt[6]{t)}