What is the solution for y^'=8xe^{(5y)},y(0)=0 ?

The solution for y^'=8xe^{(5y)},y(0)=0 is y=-(ln(-20x^2+1))/5

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y^'=8xe^{(5y)},y(0)=0

y^{^{'}} = 8 x e^{(5 y)}, y (0) = 0

y = - \frac{ln ( - 20 x ^{2} + 1 )}{5}

Solution steps

y^{'} (x) = 8 x e^{(5 y)}

Solve separable ODE:

y = - \frac{ln ( - 20 x ^{2} + 1 )}{5}

y = - \frac{ln ( - 20 x ^{2} + 1 )}{5}

\frac{d}{d x} (arccot (2 x))

\int \frac{7 ( x ^{2} + 5 )}{x ( x ^{2} + 7 )} d x

3 x^{3} - x^{2} - 10 x, - x^{2} + 2 x, [- 2, 2]

\int_{- 2}^{0} 2 x^{2} + 5 x + 6 d x

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

What is the solution for y^'=8xe^{(5y)},y(0)=0 ?
The solution for y^'=8xe^{(5y)},y(0)=0 is y=-(ln(-20x^2+1))/5