What is the solution for ty^'=3y+t^4sin(4t) ?

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

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ty^'=3y+t^4sin(4t)

t y^{^{'}} = 3 y + t^{4} sin (4 t)

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

Solution steps

t y^{'} (t) = 3 y + t^{4} sin (4 t)

Solve linear ODE:

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

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

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