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 = (5 t - 1) (6 t - 3)^{- 1}

f (x) = x + 4 e^{x}

\frac{d}{d t} (a cos (wt))

y = 2 x^{2} + 8 x + 10

\frac{d}{d x} ((4 x^{2} + 5) (5 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}