What is the solution for x^2y^'-4xy=sin(2x)x^6 ?

The solution for x^2y^'-4xy=sin(2x)x^6 is y=-1/2 cos(2x)x^4+c_{1}x^4

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x^2y^'-4xy=sin(2x)x^6

x^{2} y^{^{'}} - 4 x y = sin (2 x) x^{6}

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

Solution steps

x^{2} y^{'} (x) - 4 x y = sin (2 x) x^{6}

Solve linear ODE:

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

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

\frac{d}{d x} (a + b x)

\frac{d}{d x} (\frac{2 x e ^{x}}{x ^{2} + 1})

\int 7 x^{2} sin (2 x) d x

k = 1 \sum \infty (\frac{1}{2})^{k}

\frac{d}{d x} (\frac{x ^{2}}{64} + \frac{y ^{2}}{25}) = 1

What is the solution for x^2y^'-4xy=sin(2x)x^6 ?
The solution for x^2y^'-4xy=sin(2x)x^6 is y=-1/2 cos(2x)x^4+c_{1}x^4