What is the solution for x^2(x+1)(2x^2-x-5)dx= 1/((y+1))dy ?

The solution for x^2(x+1)(2x^2-x-5)dx= 1/((y+1))dy is y=e^{(x^6-5x^3)/3+(x^5)/5-(3x^4)/2+c_{1}}-1

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x^2(x+1)(2x^2-x-5)dx= 1/((y+1))dy

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

y = e^{\frac{x ^{6} - 5 x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{3 x ^{4}}{2} + c_{1}} - 1

Solution steps

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

Solve separable ODE:

y = e^{\frac{x ^{6} - 5 x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{3 x ^{4}}{2} + c_{1}} - 1

y = e^{\frac{x ^{6} - 5 x ^{3}}{3} + \frac{x ^{5}}{5} - \frac{3 x ^{4}}{2} + c_{1}} - 1

What is the solution for x^2(x+1)(2x^2-x-5)dx= 1/((y+1))dy ?
The solution for x^2(x+1)(2x^2-x-5)dx= 1/((y+1))dy is y=e^{(x^6-5x^3)/3+(x^5)/5-(3x^4)/2+c_{1}}-1