What is the solution for x^2y^'-3xy=sin(2x)x^5 ?

The solution for x^2y^'-3xy=sin(2x)x^5 is y=-1/2 cos(2x)x^3+c_{1}x^3

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x^2y^'-3xy=sin(2x)x^5

x^{2} y^{^{'}} - 3 x y = sin (2 x) x^{5}

y = - \frac{1}{2} cos (2 x) x^{3} + c_{1} x^{3}

Solution steps

x^{2} y^{'} (x) - 3 x y = sin (2 x) x^{5}

Solve linear ODE:

y = - \frac{1}{2} cos (2 x) x^{3} + c_{1} x^{3}

y = - \frac{1}{2} cos (2 x) x^{3} + c_{1} x^{3}

x \to 2 + lim (\frac{x ^{2}}{4 - x ^{2}})

\int \frac{e ^{x}}{1 + 16 e ^{2 x}} d x

3 x + 1

(y^{2} + y x) d x - (x^{2}) d y = 0

What is the solution for x^2y^'-3xy=sin(2x)x^5 ?
The solution for x^2y^'-3xy=sin(2x)x^5 is y=-1/2 cos(2x)x^3+c_{1}x^3