What is the solution for t(d^2y)/(dt^2)+4(dy)/(dt)=t^2 ?

The solution for t(d^2y)/(dt^2)+4(dy)/(dt)=t^2 is y=c_{1}+(c_{2})/(t^3)+(t^3)/(18)

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t(d^2y)/(dt^2)+4(dy)/(dt)=t^2

t \frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} = t^{2}

y = c_{1} + \frac{c _{2}}{t ^{3}} + \frac{t ^{3}}{18}

Solution steps

t \frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} = t^{2}

Solve linear ODE:

y = c_{1} + \frac{c _{2}}{t ^{3}} + \frac{t ^{3}}{18}

y = c_{1} + \frac{c _{2}}{t ^{3}} + \frac{t ^{3}}{18}

y = - \frac{3}{2} x^{2} + 9 x + 9, (- 1, - \frac{3}{2})

\int 7 - x^{2} d x

x \to - 2 lim (\frac{2 x}{2 x + 2})

\frac{d}{d x} ((1 + (1 + (1 + x^{3})^{4})^{5})^{6})

2 t x^{^{''}} - x^{^{'}} + (\frac{1}{x ^{^{'}}}) = 0

What is the solution for t(d^2y)/(dt^2)+4(dy)/(dt)=t^2 ?
The solution for t(d^2y)/(dt^2)+4(dy)/(dt)=t^2 is y=c_{1}+(c_{2})/(t^3)+(t^3)/(18)