What is the solution for x^2*y^'-4xy=sin(3x)*x^6 ?

The solution for x^2*y^'-4xy=sin(3x)*x^6 is y=-1/3 cos(3x)x^4+c_{1}x^4

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x^2y^'-4xy=sin(3x)x^6

x^{2} \cdot y^{^{'}} - 4 x y = sin (3 x) \cdot x^{6}

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

Solution steps

x^{2} \cdot y^{'} (x) - 4 x y = sin (3 x) \cdot x^{6}

Solve linear ODE:

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

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

What is the solution for x^2*y^'-4xy=sin(3x)*x^6 ?
The solution for x^2*y^'-4xy=sin(3x)*x^6 is y=-1/3 cos(3x)x^4+c_{1}x^4