What is the solution for x^2y^'-xy=x^4cos(2x),y(pi)= pi/4 ?

The solution for x^2y^'-xy=x^4cos(2x),y(pi)= pi/4 is y=(x(2xsin(2x)+cos(2x)))/4

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x^2y^'-xy=x^4cos(2x),y(pi)= pi/4

x^{2} y^{^{'}} - x y = x^{4} cos (2 x), y (π) = \frac{π}{4}

y = \frac{x ( 2 x sin ( 2 x ) + cos ( 2 x ) )}{4}

Solution steps

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

Solve linear ODE:

y = \frac{x ( 2 x sin ( 2 x ) + cos ( 2 x ) )}{4}

y = \frac{x ( 2 x sin ( 2 x ) + cos ( 2 x ) )}{4}

\int \frac{6 x}{( 5 + 3 x ^{2} ) ^{4}} d x

x, 5 x, - x + 4

\frac{d y}{d x}, y = x^{9}

f (x) = x^{2} - 3 x + 7

\frac{d}{d x} (x^{4} - x)

What is the solution for x^2y^'-xy=x^4cos(2x),y(pi)= pi/4 ?
The solution for x^2y^'-xy=x^4cos(2x),y(pi)= pi/4 is y=(x(2xsin(2x)+cos(2x)))/4