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

The solution for y^'=y^4*(4x^2+x) is y=-\sqrt[3]{2/(8x^3+3x^2+c_{1)}}

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

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

y = - 3 \frac{2}{8 x ^{3} + 3 x ^{2} + c _{1}}

Solution steps

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

Solve separable ODE:

y = - 3 \frac{2}{8 x ^{3} + 3 x ^{2} + c _{1}}

y = - 3 \frac{2}{8 x ^{3} + 3 x ^{2} + c _{1}}

(7, 4), (1, - 2)

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

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

5 t sin (4 t)

\int \frac{7 x ^{2} - 4}{x ^{3} + x ^{2} - 2 x} d x

What is the solution for y^'=y^4*(4x^2+x) ?
The solution for y^'=y^4*(4x^2+x) is y=-\sqrt[3]{2/(8x^3+3x^2+c_{1)}}