What is the solution for (dy)/(dx)= y/(x+1)+4x^2+4x ?

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

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(dy)/(dx)= y/(x+1)+4x^2+4x

\frac{d y}{d x} = \frac{y}{x + 1} + 4 x^{2} + 4 x

y = 2 x^{2} (x + 1) - 2 (x + 1) + c_{1} (x + 1)

Solution steps

\frac{d y}{d x} = \frac{y}{x + 1} + 4 x^{2} + 4 x

Solve linear ODE:

y = 2 x^{2} (x + 1) - 2 (x + 1) + c_{1} (x + 1)

y = 2 x^{2} (x + 1) - 2 (x + 1) + c_{1} (x + 1)

\int_{\frac{4}{3}}^{4} \frac{1}{1 - x ^{2}} d x

\int (x^{3} + 1)^{2} (3 x) d x

y = \frac{x}{x + 9}, (1, 0.1)

- 2 x^{\frac{2}{3}} - 4, at x = 8

(2 x^{2} - 4)^{3}

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