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}

x \to 2 lim (x^{3} + 5 x^{2} - 7 x + 1)

\frac{d}{d x} (- \frac{25}{( 1 + 5 x ) ^{2}})

\frac{x}{5} + \frac{5}{x}

\int 6 x^{3} 3 x^{4} + 2 d x

\int x ln (3 + x) d x

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)