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

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

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

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

y = - 3 Ci (4 t) t^{2} + c_{1} t^{2}

Solution steps

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

Solve linear ODE:

y = - 3 Ci (4 t) t^{2} + c_{1} t^{2}

y = - 3 Ci (4 t) t^{2} + c_{1} t^{2}

\frac{\partial}{\partial x} (\frac{1}{2} e^{2 x})

f (x) = x^{[2]} + 1, at x = 3

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

n = 1 \sum \infty (\frac{1}{11})^{n}

\int x \cdot sin (a x) d x

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