What is the solution for ty^'+4y=t^3,y(2)=4 ?

The solution for ty^'+4y=t^3,y(2)=4 is y=(t^3)/7+(320)/(7t^4)

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ty^'+4y=t^3,y(2)=4

t y^{^{'}} + 4 y = t^{3}, y (2) = 4

y = \frac{t ^{3}}{7} + \frac{320}{7 t ^{4}}

Solution steps

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

Solve linear ODE:

y = \frac{t ^{3}}{7} + \frac{320}{7 t ^{4}}

y = \frac{t ^{3}}{7} + \frac{320}{7 t ^{4}}

\frac{d}{d x} (\frac{w}{1 - 4 x ^{2}})

2 t^{- \frac{1}{4}} + 2 ln (t) + ln (5)

y^{^{''}} - y^{^{'}} + 0.25 y = 0

\int e^{4 x} sin (b x) d x

f (x) = x + 9

What is the solution for ty^'+4y=t^3,y(2)=4 ?
The solution for ty^'+4y=t^3,y(2)=4 is y=(t^3)/7+(320)/(7t^4)