Question

1. ( 12pts) solve the differential equation $yy(x + 1)=y^{2}+15$

Explanation:

Step1: Separate variables

Given \(y'y(x + 1)=y^{2}+15\), rewrite it as \(\frac{y}{y^{2}+15}dy=\frac{1}{x + 1}dx\)

Step2: Integrate both sides

Integrate \(\int\frac{y}{y^{2}+15}dy\) and \(\int\frac{1}{x + 1}dx\)
For \(\int\frac{y}{y^{2}+15}dy\), let \(u = y^{2}+15\), then \(du=2ydy\) and \(\int\frac{y}{y^{2}+15}dy=\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u|+C_1=\frac{1}{2}\ln(y^{2}+15)+C_1\)
For \(\int\frac{1}{x + 1}dx=\ln|x + 1|+C_2\)
So \(\frac{1}{2}\ln(y^{2}+15)=\ln|x + 1|+C\) (where \(C = C_2 - C_1\))

Step3: Simplify the equation

Multiply both sides by \(2\): \(\ln(y^{2}+15)=2\ln|x + 1|+2C\)
Using the property \(a\ln b=\ln(b^{a})\), we have \(\ln(y^{2}+15)=\ln((x + 1)^{2})+\ln K\) (where \(K = e^{2C}\))
By the property \(\ln a+\ln b=\ln(ab)\), we get \(y^{2}+15 = K(x + 1)^{2}\)
Then \(y^{2}=K(x + 1)^{2}-15\)
\(y=\pm\sqrt{K(x + 1)^{2}-15}\)

Answer:

\(y=\pm\sqrt{K(x + 1)^{2}-15}\), where \(K\) is an arbitrary constant.