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

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

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x^2y^'+(2x^2-x)y=x^4

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

y = \frac{x ^{2}}{2} - \frac{x}{4} + \frac{c _{1} x}{e ^{2 x + c_{1}}}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{2}}{2} - \frac{x}{4} + \frac{c _{1} x}{e ^{2 x + c_{1}}}

y = \frac{x ^{2}}{2} - \frac{x}{4} + \frac{c _{1} x}{e ^{2 x + c_{1}}}

\frac{\partial}{\partial y} (x^{2} - \frac{y}{4})

9

f (x) = \frac{8 x}{x ^{2} + 1}, at x = 0

\frac{d}{d x} (x (1 - 2 y))

\frac{25 x ^{3}}{3} + 10 x^{2} + 4 x - 7

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