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

The solution for y^'=((x^2+y))/(x+1) is y=x(x+1)-2ln(x+1)(x+1)-1+c_{1}(x+1)

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y^'=((x^2+y))/(x+1)

y^{^{'}} = \frac{( x ^{2} + y )}{x + 1}

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

Solution steps

y^{'} (x) = \frac{( x ^{2} + y )}{x + 1}

Solve linear ODE:

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

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

y^{^{'}} + \frac{y}{x} = x^{- 4}

s (t) = 14 + 15 sin (14 π t)

(x^{2} - 2 x y) d x + x d y = 0

sin^{2} (π t)

\int sin (\frac{π}{2}) d x

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