What is the solution for y^{''}+(2/x)y^'=2x+4 ?

The solution for y^{''}+(2/x)y^'=2x+4 is y=c_{1}+(c_{2})/x+(x^3+4x^2)/6

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y^{''}+(2/x)y^'=2x+4

y^{^{''}} + (\frac{2}{x}) y^{^{'}} = 2 x + 4

y = c_{1} + \frac{c _{2}}{x} + \frac{x ^{3} + 4 x ^{2}}{6}

Solution steps

y^{''} (x) + (\frac{2}{x}) y^{'} (x) = 2 x + 4

Solve linear ODE:

y = c_{1} + \frac{c _{2}}{x} + \frac{x ^{3} + 4 x ^{2}}{6}

y = c_{1} + \frac{c _{2}}{x} + \frac{x ^{3} + 4 x ^{2}}{6}

(x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0, y (1) = 0

\frac{d}{d x} (arccsc (x))

\frac{2 s + 1}{s ^{2} + 2 s + 1}

\frac{d}{d x} (m e^{m x})

x \to 0 lim (x 1 + \frac{1}{2})

What is the solution for y^{''}+(2/x)y^'=2x+4 ?
The solution for y^{''}+(2/x)y^'=2x+4 is y=c_{1}+(c_{2})/x+(x^3+4x^2)/6