What is the solution for ty^{''}+4y^'=t^2 ?

The solution for ty^{''}+4y^'=t^2 is y=c_{1}+(c_{2})/(t^3)+(t^3)/(18)

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ty^{''}+4y^'=t^2

t y^{^{''}} + 4 y^{^{'}} = t^{2}

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

Solution steps

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

f f^{^{'}} (xx) = x x^{3} - x x^{2} - 6 xx

\frac{d}{d x} (e^{c o s h (x)} cos^{2} (x))

\int_{0}^{7} - 30 x^{2} + 420 x d x

\int \frac{10 - 10 x}{1 - x} d x

\int \frac{x}{6 x - x ^{2}} d x

What is the solution for ty^{''}+4y^'=t^2 ?
The solution for ty^{''}+4y^'=t^2 is y=c_{1}+(c_{2})/(t^3)+(t^3)/(18)