Problem 69
Find the integral curve of xy'-(y)/(x+1)=x that passes…
Find the integral curve of $xy'-\dfrac{y}{x+1}=x$ that passes through the point $(1,-1)$.
Explanation of Method
Divide by $x$ to get standard linear form $y'-\dfrac{1}{x(x+1)}y=1$, then apply the integrating factor found from partial fractions.
Step 1: Find the integrating factor
Since $\dfrac{1}{x(x+1)}=\dfrac1x-\dfrac1{x+1}$, $\displaystyle\int -\frac{dx}{x(x+1)}=-\ln|x|+\ln|x+1|$, so $\mu(x)=\dfrac{x+1}{x}$.
Step 2: Multiply and integrate
$\dfrac{d}{dx}\!\left[\dfrac{(x+1)y}{x}\right]=\dfrac{x+1}{x}=1+\dfrac1x$, so $\dfrac{(x+1)y}{x}=x+\ln|x|+C$.
Step 3: General solution and initial condition
Hence $y=\dfrac{x^2+x\ln|x|+Cx}{x+1}$. At $(1,-1)$: $-1=\dfrac{1+0+C}{2}\Rightarrow C=-3$.
Result
Final answer:
General solution: \[y=\dfrac{x^2+x\ln|x|+Cx}{x+1},\qquad C\in\mathbb{R}\]
Particular solution through $(1,-1)$:
\[\boxed{\,y=\dfrac{x^2-3x+x\ln x}{x+1}.}\]