What is the solution for 13(t+1)(dy)/(dt)-9y=36t ?

The solution for 13(t+1)(dy)/(dt)-9y=36t is y=9t+13+c_{1}(t+1)^{9/13}

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13(t+1)(dy)/(dt)-9y=36t

13 (t + 1) \frac{d y}{d t} - 9 y = 36 t

y = 9 t + 13 + c_{1} (t + 1)^{\frac{9}{13}}

Solution steps

13 (t + 1) \frac{d y}{d t} - 9 y = 36 t

Solve linear ODE:

y = 9 t + 13 + c_{1} (t + 1)^{\frac{9}{13}}

y = 9 t + 13 + c_{1} (t + 1)^{\frac{9}{13}}

\int x cos (10 x) d x

\frac{d}{d x} (arcsec (\frac{1}{x}))

\int_{- 1}^{1} (1 - x) d x

\int \frac{x - 1}{x ^{2} + 9} d x

y^{^{''}} + 4 y^{^{'}} + 5 y = 8

What is the solution for 13(t+1)(dy)/(dt)-9y=36t ?
The solution for 13(t+1)(dy)/(dt)-9y=36t is y=9t+13+c_{1}(t+1)^{9/13}