Question

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

First order differential equations
Solve differential equation $$xu'(x)= u^2-4$$

2021-03-08

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