What is the solution for y^'+3/t y=1-2t,y(3)=0 ?

The solution for y^'+3/t y=1-2t,y(3)=0 is y= t/4-(2t^2)/5+(1539)/(20t^3)

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y^'+3/t y=1-2t,y(3)=0

y^{^{'}} + \frac{3}{t} y = 1 - 2 t, y (3) = 0

y = \frac{t}{4} - \frac{2 t ^{2}}{5} + \frac{1539}{20 t ^{3}}

Solution steps

y^{'} (t) + \frac{3}{t} y = 1 - 2 t

Solve linear ODE:

y = \frac{t}{4} - \frac{2 t ^{2}}{5} + \frac{1539}{20 t ^{3}}

y = \frac{t}{4} - \frac{2 t ^{2}}{5} + \frac{1539}{20 t ^{3}}

\frac{d}{d x} (sin (2 x + y))

\int_{0}^{2 π} sin (x) cos (x) d x

y^{^{''}} - 3 y^{^{'}} = 4 t

\int 2 e^{7} d x

\int_{- 3}^{- 2} \frac{x + 2}{x ^{2} + 4 x} d x

What is the solution for y^'+3/t y=1-2t,y(3)=0 ?
The solution for y^'+3/t y=1-2t,y(3)=0 is y= t/4-(2t^2)/5+(1539)/(20t^3)