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I'm given the ODE:

$\left(4+x\right)y'+\frac{1}{2}y=\frac{4+x}{\sqrt{4-x}}$

I try to solve it regularly and I get $y(x) = \sqrt{4+x}(\ln|4+x|+C)$ for some constant $C$. I'm also not entirely sure I got the math correctly, however, the solution has two answers:

$y(x) = \bigg(\sin^{-1}(\frac{x}{4}) +a\bigg)\sqrt{4+x}$

if $x > -4$, and

$y(x) = \bigg(\ln|x+\sqrt{x^2 -16}| + a\bigg) \sqrt{|x+4|}$

if $x \leq -4$.

Looking at these 2 solutions, it looks like my original answer is incorrect, but I'm unsure why there are two different answers for 2 intervals. I'm guessing because there is a discontinuity in the denominator in the right-hand side, but I don't know how to deal with it. Can someone explain this?

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Hint:

without RHS

$\frac{y'}{y}=\frac{-1}{2(x+4)}$

which gives

$$ln(\frac{y}{\lambda})=\frac{-1}{2}ln(|x+4|)$$

and

$$y=\frac{\lambda}{\sqrt{|x+4|}}$$

now, let us look for a particular solution of the form

$$y_p=\frac{\lambda(x)}{\sqrt{|x+4|}}$$

if we replace, we get

$$(4+x)\frac{\lambda'(x)}{\sqrt{|x+4|}}= \frac{x+4}{\sqrt{4-x}}$$

or

$$\lambda'(x)=\sqrt{\frac{|x+4|}{4-x}}$$

which could be integrated at$(-\infty,-4]$ and at $[-4,4)$.

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