What is the solution for 2ty'+(4+12t^3-120t^2)y=0 ?

The solution for 2ty'+(4+12t^3-120t^2)y=0 is y=(c_{1})/(e^{2t^3-30t^2)t^2}

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2ty'+(4+12t^3-120t^2)y=0

2 t y^{^{'}} + (4 + 12 t^{3} - 120 t^{2}) y = 0

y = \frac{c _{1}}{e ^{2 t^{3} - 30 t^{2}} t ^{2}}

Solution steps

2 t y^{'} (t) + (4 + 12 t^{3} - 120 t^{2}) y = 0

Solve linear ODE:

y = \frac{c _{1}}{e ^{2 t^{3} - 30 t^{2}} t ^{2}}

y = \frac{c _{1}}{e ^{2 t^{3} - 30 t^{2}} t ^{2}}

\frac{6 s + 3}{s ^{2}}

\int \frac{x}{( x ^{2} - 7 x + 12 )} d x

n = 0 \sum \infty (\frac{x}{2} - 3)^{n}

\frac{d}{d x} ((2 r (x) a^{r (x) x} + n)^{p (x)})

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

What is the solution for 2ty'+(4+12t^3-120t^2)y=0 ?
The solution for 2ty'+(4+12t^3-120t^2)y=0 is y=(c_{1})/(e^{2t^3-30t^2)t^2}