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

\frac{( x ^{2} - 4 ) 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

\frac{( x ^{2} - 4 ) 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^{2} + 5, x^{2} - 10 x + 17, 2, 3

f (x) = \frac{1}{( x ^{6} - \frac{7}{x} ) ^{2}}

- \frac{4}{( 4 x - 17 ) ^{\frac{3}{2}}}

\frac{\partial}{\partial x} (ln (4 x + y))

n = 0 \sum \infty e^{- 2 nab}

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)