Linear Equations — Question 1

Question 1 – Solution

Strategy. Normalize first. Choose $\mu$ with $\mu'=p\mu$, so the product rule gives $(\mu y)'=\mu(y'+py)$.

Step 1: Integrating factor. Since $1+x^2>0$, $$y'+\frac{2x}{1+x^2}y=\frac{3x^2+1}{1+x^2},\qquad \mu=e^{\int 2x/(1+x^2)\,dx}=1+x^2.$$ Multiplication restores the original left side as $[(1+x^2)y]'$.

Step 2: Integrate and select the constant. We obtain $$(1+x^2)y=\int(3x^2+1)\,dx=x^3+x+C,$$ so $$\boxed{y=x+\frac{C}{1+x^2}.}$$ At $x=0$, $C=2$. The IVP solution is therefore $\boxed{y=x+2/(1+x^2)}$.

Step 3: Direct check and interval. Its derivative is $y'=1-4x/(1+x^2)^2$, and $$\begin{aligned} (1+x^2)y'+2xy &=1+x^2-\frac{4x}{1+x^2}+2x^2+\frac{4x}{1+x^2}\\ &=3x^2+1. \end{aligned}$$ Also $y(0)=2$. No denominator vanishes for real $x$, so the largest open interval is $\boxed{\mathbb R}$.

Step 4: Approach to a line. Setting $C=0$ gives the particular solution $y_p=x$. For every fixed real $C$, $$\boxed{y(x)-x=\frac{C}{1+x^2}\longrightarrow 0\quad\text{as }x\to\pm\infty.}$$ It is the vertical difference from the line that tends to zero; the solution itself does not tend to a finite constant.