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

The solution for y^'-2/t y=t^2e^t,y(1)=3 is y=e^tt^2+(3-e)t^2

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

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

y = e^{t} t^{2} + (3 - e) t^{2}

Solution steps

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

Solve linear ODE:

y = e^{t} t^{2} + (3 - e) t^{2}

y = e^{t} t^{2} + (3 - e) t^{2}

5^{x^{2}}

(1 - x^{3}) y^{^{'}} - 2 (1 + x) y = y^{\frac{5}{2}}

\int 8 + 6 x + 24 x^{2} d x

\frac{d}{d x} (\frac{9 e ^{x} x ^{\frac{7}{2}}}{2} + e^{x} x^{\frac{9}{2}})

\frac{\partial}{\partial x} (\frac{4 x + 9}{3 y + 1})

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