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}}

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

\frac{\partial}{\partial y} (7 sin (x y))

\frac{\partial}{\partial x} (3 x e^{5 x y})

x \to - 4 lim (x)

\int \frac{x ^{2}}{( 1 + 2 x ) ^{\frac{1}{3}}} d x

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}