What is the solution for 2xy^'-y=x^2 ?

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

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2xy^'-y=x^2

2 x y^{^{'}} - y = x^{2}

y = \frac{1}{3} x^{\frac{3}{2}} x + c_{1} x

Solution steps

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

Solve linear ODE:

y = \frac{1}{3} x^{\frac{3}{2}} x + c_{1} x

y = \frac{1}{3} x^{\frac{3}{2}} x + c_{1} x

\int_{6}^{8} \frac{x}{x ^{2} + 4 x + 13} d x

\frac{d}{d x} (x \cdot ln (x) + (3 \cdot x^{2} - 2)^{3})

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int 14 x^{\frac{3}{2}} d x

\int \frac{y ^{2}}{3} d y

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