What is the solution for 8xy^'=2xe^x-8y+6x^2 ?

The solution for 8xy^'=2xe^x-8y+6x^2 is y=(e^xx-e^x}{4x}+\frac{x^2)/4+(c_{1})/x

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8xy^'=2xe^x-8y+6x^2

8 x y^{^{'}} = 2 x e^{x} - 8 y + 6 x^{2}

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

Solution steps

8 x y^{'} (x) = 2 x e^{x} - 8 y + 6 x^{2}

Solve linear ODE:

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

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

h \to 0 lim (\frac{( - 5 + h ) - 25}{h})

x \to \infty lim ((i)^{x})

\int \frac{x ^{2}}{( 2 - x ^{3} ) ^{2}} d x

\frac{d}{d x} (x^{2} + 64)

\frac{\partial}{\partial z} (- 3 e^{x} cos (yz))

What is the solution for 8xy^'=2xe^x-8y+6x^2 ?
The solution for 8xy^'=2xe^x-8y+6x^2 is y=(e^xx-e^x}{4x}+\frac{x^2)/4+(c_{1})/x