# Solve differential equation xu'(x)= u^2-4

Question
Solve differential equation $$xu'(x)= u^2-4$$

2021-03-08
$$xu'(x)= u^2-4$$
$$xu'(x) 1/(u^2-4)=1$$
$$1/(u^2-4) du=1/x dx$$
Integrating on both sides
$$int 1/(u^2-4)du= int 1/x dx$$
The partial fraction of the left side $$1/((u+2)(u−2))= −1/(4(x+2))+1/(4(x−2))$$
$$-1/4 int 1/(u+2)du +1/4 int 1/(u-2)du= int 1/x dx$$
$$-1/4 ln (u+2)du+1/4 ln(u-2)= ln(x)+C_1$$
$$-ln(u+2)du+ln(u-2)= 4 ln(x)+4C_1$$
$$ln((u-2)/(u+2))= ln(x^4)+C_2$$
$$((u-2)/(u+2))= (x^4)+e^(C_2)$$
$$= x^4+C$$
$$u= - (2(C+x^4))/(-C+x^4)$$

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