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

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

x \to 0 lim (\frac{( cos ( 2 x ) - cos ( 3 x ) )}{x ^{2}})

\int \frac{( x + 3 )}{5 - 4 x - x ^{2}} d x

\frac{\partial}{\partial x} (sin (3 y) + 2 z e^{- 2 x})

\frac{\partial}{\partial x} (x - \frac{1}{y})

\int (2 x + 3) 2 x + 1 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=(2ln(x+2)(x+2))/(x-2)+(c_{1}(x+2))/(x-2)