What is the solution for (x+2)^2y^'+3(x+2)y=4 ?

The solution for (x+2)^2y^'+3(x+2)y=4 is y=(2(x^2+4x))/(x^3+6x^2+12x+8)+(c_{1})/(x^3+6x^2+12x+8)

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(x+2)^2y^'+3(x+2)y=4

(x + 2)^{2} y^{^{'}} + 3 (x + 2) y = 4

y = \frac{2 ( x ^{2} + 4 x )}{x ^{3} + 6 x ^{2} + 12 x + 8} + \frac{c _{1}}{x ^{3} + 6 x ^{2} + 12 x + 8}

Solution steps

(x + 2)^{2} y^{'} (x) + 3 (x + 2) y = 4

Solve linear ODE:

y = \frac{2 ( x ^{2} + 4 x )}{x ^{3} + 6 x ^{2} + 12 x + 8} + \frac{c _{1}}{x ^{3} + 6 x ^{2} + 12 x + 8}

y = \frac{2 ( x ^{2} + 4 x )}{x ^{3} + 6 x ^{2} + 12 x + 8} + \frac{c _{1}}{x ^{3} + 6 x ^{2} + 12 x + 8}

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

\frac{d y}{d x} = 2 y + y^{5}, y (0) = 1

\int e^{6 t} d t

f (x) = \frac{A}{x ^{10}} + B e^{x}

y^{^{'}} - 5 y = 6 e^{5 x}, y (0) = 0

What is the solution for (x+2)^2y^'+3(x+2)y=4 ?
The solution for (x+2)^2y^'+3(x+2)y=4 is y=(2(x^2+4x))/(x^3+6x^2+12x+8)+(c_{1})/(x^3+6x^2+12x+8)