What is the solution for (x+1)y^'-3y=0 ?

The solution for (x+1)y^'-3y=0 is y=e^{c_{1}}x^3+3e^{c_{1}}x^2+3e^{c_{1}}x+e^{c_{1}}

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(x+1)y^'-3y=0

(x + 1) y^{^{'}} - 3 y = 0

y = e^{c_{1}} x^{3} + 3 e^{c_{1}} x^{2} + 3 e^{c_{1}} x + e^{c_{1}}

Solution steps

(x + 1) y^{'} (x) - 3 y = 0

Solve separable ODE:

y = e^{c_{1}} x^{3} + 3 e^{c_{1}} x^{2} + 3 e^{c_{1}} x + e^{c_{1}}

y = e^{c_{1}} x^{3} + 3 e^{c_{1}} x^{2} + 3 e^{c_{1}} x + e^{c_{1}}

y^{^{'}} = 3 x^{2} y^{2}, y (- 1) = a, y (0) = 1

\frac{d y}{d x} = \frac{x y - 4 x + y - 4}{x y - 6 x - 3 y + 18}

\frac{d y}{d x} = x^{2} (y - 2), y (0) = 4

y^{^{'}} = e^{- y}, y (0) = 0

\frac{d y}{d x} = \frac{2}{3} x 1 - 9 y^{2}, y (0) = 0

What is the solution for (x+1)y^'-3y=0 ?
The solution for (x+1)y^'-3y=0 is y=e^{c_{1}}x^3+3e^{c_{1}}x^2+3e^{c_{1}}x+e^{c_{1}}