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

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

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

\frac{d y}{d x} = \frac{1}{x + 1} y + 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{1}{x + 1} y + 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)

2 (t - 2)^{2}

\frac{d y}{d x} = \frac{3 x + 1}{4 y}

\int_{0}^{1} \frac{5 x ^{4}}{1 + x ^{5}} d x

\int (6 x - 2)^{- 4} d x

\int \frac{2 r}{( 1 - r ) ^{7}} d r

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