What is the solution for x^2y^'-2xy=sin(4x)x^4 ?

The solution for x^2y^'-2xy=sin(4x)x^4 is y=-1/4 cos(4x)x^2+c_{1}x^2

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x^2y^'-2xy=sin(4x)x^4

x^{2} y^{^{'}} - 2 x y = sin (4 x) x^{4}

y = - \frac{1}{4} cos (4 x) x^{2} + c_{1} x^{2}

Solution steps

x^{2} y^{'} (x) - 2 x y = sin (4 x) x^{4}

Solve linear ODE:

y = - \frac{1}{4} cos (4 x) x^{2} + c_{1} x^{2}

y = - \frac{1}{4} cos (4 x) x^{2} + c_{1} x^{2}

arctan (3 x)

\frac{d}{d t} (sin^{2} (t))

\frac{d}{d x} (6 e^{x} + 2 y)

(2 x - 4) \frac{d y}{d x} - y^{2} = 0

\frac{d}{d x} (cos (x) cos (x))

What is the solution for x^2y^'-2xy=sin(4x)x^4 ?
The solution for x^2y^'-2xy=sin(4x)x^4 is y=-1/4 cos(4x)x^2+c_{1}x^2