 Laila Murphy

2022-11-18

Find the solution to the differential equation
Assume $x>0$ and let
$x\left(x+1\right)\frac{du}{dx}={u}^{2},$
$u\left(1\right)=4.$
I started off by doing some algebra to get:
$\frac{1}{{u}^{2}}du=\frac{1}{{x}^{2}+x}dx.$
I then took the partial fraction of the right side of the equation:
$\frac{1}{{u}^{2}}du=\left(\frac{1}{x}-\frac{1}{x+1}\right).$
I then took the integral of both sides:
$-\frac{1}{u}=\mathrm{log}x-\mathrm{log}\left(x+1\right)+C.$
From here I don't know what to do because we are solving for $u\left(x\right)$ and I'm not sure how to get that from $-\frac{1}{u}$ grizintimbp

Expert

Solving for $\frac{du}{dx}$ we have
$\frac{du}{dx}=\frac{{u}^{2}}{x\left(x+1\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{du}{dx}=\frac{{u}^{2}}{{x}^{2}+x}\phantom{\rule{0ex}{0ex}}\frac{\frac{du}{dx}}{{u}^{2}}=\frac{1}{{x}^{2}+x}.$
Integrate both sides & evaluate the integrals:
$-\frac{1}{u}=\mathrm{log}\left(x\right)-\mathrm{log}\left(x+1\right)+{C}_{1}\phantom{\rule{0ex}{0ex}}⇒u=-\frac{1}{\mathrm{log}\left(x\right)-\mathrm{log}\left(x+1\right)+{C}_{1}}.$
Now apply the initial condition:
$-\frac{1}{{C}_{1}-\mathrm{log}\left(2\right)}=4⇒{C}_{1}=\frac{1}{4}\left(-1+4\mathrm{log}\left(2\right)\right).$
This gives the result.

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