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

The solution for 2ty^'-4y=4t^4e^{t^2} is y=e^{t^2}t^2+c_{1}t^2

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2ty^'-4y=4t^4e^{t^2}

2 t y^{^{'}} - 4 y = 4 t^{4} e^{t^{2}}

y = e^{t^{2}} t^{2} + c_{1} t^{2}

Solution steps

2 t y^{'} (t) - 4 y = 4 t^{4} e^{t^{2}}

Solve linear ODE:

y = e^{t^{2}} t^{2} + c_{1} t^{2}

y = e^{t^{2}} t^{2} + c_{1} t^{2}

\frac{\partial}{\partial x} (4 e^{5 x y})

x \to + 0 lim (\frac{1}{csc ( x )})

\int \frac{1}{100 x ^{2} + 1} d x

y^{^{'}} = \frac{y}{x} + \frac{6 x ^{3}}{y} cos (x^{2})

y = 2 x + x^{2}, (- 2, 0)

What is the solution for 2ty^'-4y=4t^4e^{t^2} ?
The solution for 2ty^'-4y=4t^4e^{t^2} is y=e^{t^2}t^2+c_{1}t^2