What is the solution for (x^2-4)((dy)/(dx))+4y=(x+2)^2 ?

The solution for (x^2-4)((dy)/(dx))+4y=(x+2)^2 is y=(x(x+2))/(x-2)+(c_{1}(x+2))/(x-2)

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(x^2-4)((dy)/(dx))+4y=(x+2)^2

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

y = \frac{x ( x + 2 )}{x - 2} + \frac{c _{1} ( x + 2 )}{x - 2}

Solution steps

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

Solve linear ODE:

y = \frac{x ( x + 2 )}{x - 2} + \frac{c _{1} ( x + 2 )}{x - 2}

y = \frac{x ( x + 2 )}{x - 2} + \frac{c _{1} ( x + 2 )}{x - 2}

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y^{^{''}} + y = k \cdot (e^{(- a \cdot t)} - e^{(- b \cdot t)})

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What is the solution for (x^2-4)((dy)/(dx))+4y=(x+2)^2 ?
The solution for (x^2-4)((dy)/(dx))+4y=(x+2)^2 is y=(x(x+2))/(x-2)+(c_{1}(x+2))/(x-2)