What is the solution for (x+2)^2y^'+5(x+2)y=16,y(0)=4 ?

The solution for (x+2)^2y^'+5(x+2)y=16,y(0)=4 is y= 4/(x+2)+(64)/(x^5+10x^4+40x^3+80x^2+80x+32)

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(x+2)^2y^'+5(x+2)y=16,y(0)=4

(x + 2)^{2} y^{^{'}} + 5 (x + 2) y = 16, y (0) = 4

y = \frac{4}{x + 2} + \frac{64}{x ^{5} + 10 x ^{4} + 40 x ^{3} + 80 x ^{2} + 80 x + 32}

Solution steps

(x + 2)^{2} y^{'} (x) + 5 (x + 2) y = 16

Solve linear ODE:

y = \frac{4}{x + 2} + \frac{64}{x ^{5} + 10 x ^{4} + 40 x ^{3} + 80 x ^{2} + 80 x + 32}

y = \frac{4}{x + 2} + \frac{64}{x ^{5} + 10 x ^{4} + 40 x ^{3} + 80 x ^{2} + 80 x + 32}

\int 5 x^{- 7} d x

x^{3} y^{2} = x y^{3} + 6, (2, 1)

\frac{d}{d x} (lo g_{5} (6 x^{3}))

y = \frac{1}{x - 4}, at x = 9

\int 7 e^{2} d x

What is the solution for (x+2)^2y^'+5(x+2)y=16,y(0)=4 ?
The solution for (x+2)^2y^'+5(x+2)y=16,y(0)=4 is y= 4/(x+2)+(64)/(x^5+10x^4+40x^3+80x^2+80x+32)