What is the solution for -8x^2y+y^'=-5x^2 ?

The solution for -8x^2y+y^'=-5x^2 is y=(e^{(8x^3+c_{1})/3})/(8)+5/8

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-8x^2y+y^'=-5x^2

- 8 x^{2} y + y^{^{'}} = - 5 x^{2}

y = \frac{e ^{\frac{8 x ^{3} + c _{1}}{3}}}{8} + \frac{5}{8}

Solution steps

- 8 x^{2} y + y^{'} (x) = - 5 x^{2}

Solve separable ODE:

y = \frac{e ^{\frac{8 x ^{3} + c _{1}}{3}}}{8} + \frac{5}{8}

y = \frac{e ^{\frac{8 x ^{3} + c _{1}}{3}}}{8} + \frac{5}{8}

\int \frac{1}{3 - 4 sin ^{2} ( x )} d x

f (x) = x^{2020}

\int \frac{2 x ^{2}}{36 - x ^{2}} d x

\int 2 sec (x) tan (x) - 5 sec^{2} (x) d x

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

What is the solution for -8x^2y+y^'=-5x^2 ?
The solution for -8x^2y+y^'=-5x^2 is y=(e^{(8x^3+c_{1})/3})/(8)+5/8