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

The solution for (2+x)y^'=2y is y=4e^{c_{1}}+4e^{c_{1}}x+e^{c_{1}}x^2

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(2+x)y^'=2y

(2 + x) y^{^{'}} = 2 y

y = 4 e^{c_{1}} + 4 e^{c_{1}} x + e^{c_{1}} x^{2}

Solution steps

(2 + x) y^{'} (x) = 2 y

Solve separable ODE:

y = 4 e^{c_{1}} + 4 e^{c_{1}} x + e^{c_{1}} x^{2}

y = 4 e^{c_{1}} + 4 e^{c_{1}} x + e^{c_{1}} x^{2}

\frac{d x}{d t} - 0.04 x = 5000, x (0) = 18000

y^{^{'}} = \frac{y ^{2}}{x ^{2} y + x y}

\frac{d y}{d x} y = x (2 x + 1)^{\frac{1}{3}}

\frac{d y}{d x} = \frac{ln ( 2 )}{7} y - 80000

\frac{d y}{d x} = 3 \cdot \frac{y ^{5}}{x ^{4}}

What is the solution for (2+x)y^'=2y ?
The solution for (2+x)y^'=2y is y=4e^{c_{1}}+4e^{c_{1}}x+e^{c_{1}}x^2