Question

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

First order differential equations
ANSWERED
asked 2021-03-07
Solve differential equation \(xu'(x)= u^2-4\)

Answers (1)

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 \(\frac{1}{(u+2)(uāˆ’2)}= āˆ’\frac{1}{4(x+2)}+\frac{1}{4(xāˆ’2)}\)
\(-\frac{1}{4} \int \frac{1}{u+2}du +\frac{1}{4} \int \frac{1}{u-2}du= \int \frac{1}{x} dx\)
\(-\frac{1}{4} \ln (u+2)du+\frac{1}{4} \ln(u-2)= \ln(x)+C_1\)
\(-\ln(u+2)du+ln(u-2)= 4 \ln(x)+4C_1\)
\(\ln(\frac{u-2}{u+2})= \ln(x^4)+C_2\)
\((\frac{u-2}{u+2})= (x^4)+e^{C_2}\)
\(= x^4+C\)
\(u= - \frac{2(C+x^4)}{-C+x^4}\)

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