What is the solution for x^2dy-2xydx=x^4cos(2x)dx ?

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

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x^2dy-2xydx=x^4cos(2x)dx

x^{2} d y - 2 x y d x = x^{4} cos (2 x) d x

y = \frac{1}{2} sin (2 x) x^{2} + c_{1} x^{2}

Solution steps

x^{2} d y - 2 x y d x = x^{4} cos (2 x) d x

Solve linear ODE:

y = \frac{1}{2} sin (2 x) x^{2} + c_{1} x^{2}

y = \frac{1}{2} sin (2 x) x^{2} + c_{1} x^{2}

\int \frac{2 x ^{2} + 7 x - 3}{x - 2} d x

ln (arctan (x))

(t + e^{t}) (6 - t)

y^{^{''}} + 5 y^{^{'}} + 6 y = 6

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

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