What is the solution for ty^'=2y-3t^3cos(4t) ?

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

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ty^'=2y-3t^3cos(4t)

t y^{^{'}} = 2 y - 3 t^{3} cos (4 t)

y = - \frac{3}{4} sin (4 t) t^{2} + c_{1} t^{2}

Solution steps

t y^{'} (t) = 2 y - 3 t^{3} cos (4 t)

Solve linear ODE:

y = - \frac{3}{4} sin (4 t) t^{2} + c_{1} t^{2}

y = - \frac{3}{4} sin (4 t) t^{2} + c_{1} t^{2}

\int \frac{5}{2 + 5 x} d x

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

\frac{d}{d x} ((sin (5 x))^{2})

\frac{d}{d x} (\frac{1}{x ln ( x )})

x \to 0 - lim (1 + x e^{x})

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