Differential equation with really strange solution... I wish to solve the following differential equation $$\frac{dy}{dx}=\frac{\sqrt{y+7}}{(x+3)(x+8)}$$ with initial condition $y(0)=2$. I think my method is correct but my final solution seems like a really strange answer... This is what I got:
$$\Rightarrow y=2+\frac{1}{100}\left(\ln\left|\frac{3(x+3)}{8(x+8)}\right|\right)\ln\left(\left|\frac{8(x+3)}{3(x+8)}\right|\right)$$
I’m not sure how to check if this is correct; is it right? Thanks!!
This is my working:
We have the differential equation $\frac{dy}{dx}=\frac{\sqrt{(y+7)}}{(x+3)(x+8)}$. This is clearly separable, so we rearrange to get the following:
$$\frac{1}{\sqrt{y+7}}dy=\int\frac{1}{(x+8)(x+3)}dx$$
$$\Rightarrow 2\sqrt{y+7}=\frac{1}{5}\ln\left|\frac{x+3}{x+8}\right|$$
$$\Rightarrow \sqrt{y+7}=\frac{1}{10}\ln\left|\frac{x+3}{x+8}\right|$$
$$\Rightarrow y+7=\frac{1}{100}\left(\ln\left|\frac{x+3}{x+8}\right|\right)^2$$
$$\Rightarrow y=-7+\frac{1}{100}\left(\ln\left|\frac{x+3}{x+8}\right|\right)^2+C$$
Now set $y(0)=2$:
$$2=-7+\frac{1}{100}\left(\ln\left|\frac{0+3}{0+8}\right|\right)^2+C$$
$$\Rightarrow 9=\frac{1}{100}\left(\ln\left|\frac{3}{8}\right|\right)^2+C$$
$$\Rightarrow 900=\left(\ln\frac{3}{8}\right)^2+100C$$
$$\Rightarrow 100C=900-\ln\left(\frac{3}{8}\right)^2$$
$$\Rightarrow C=9-\frac{1}{100}\ln\left(\frac{3}{8}\right)^2$$
Substituting this value of $C$ into the general solution yields the following:
$$y=-7+\frac{1}{100}\left(\ln\left|\frac{x+3}{x+8}\right|\right)^2+9-\frac{1}{100}\left(\ln\frac{3}{8}\right)^2$$
$$\Rightarrow y=2+\frac{1}{100}\left[\left(\ln\frac{x+3}{x+8}\right)^2-\ln\left(\frac{3}{8}\right)^2\right]$$
$$\Rightarrow y=2+\frac{1}{100}\left[\ln\left|\frac{x+3}{x+8}\right|+\ln\left(\frac{3}{8}\right)\right]\left[\ln\left|\frac{x+3}{x+8}\right|-\ln\frac{3}{8}\right],$$ by the difference of two squares identity.
$$\Rightarrow y=2+\frac{1}{100}\left(\ln\left|\frac{3(x+3)}{8(x+8)}\right|\right)\left(\left|\frac{8(x+3)}{3(x+8)}\right|\right)$$

You should have added $C$, the constant, before squaring. That made all the difference.
