Question

1. $(1 + x^{2})\frac{dy}{dx}+2xy = f(x); y(0)=0,$

$f(x)=\begin{cases}x, & 0leq xlt1\\ -x, & xgeq1end{cases}$

Explanation:

Step1: Rewrite the ODE

The given first - order linear ODE \((1 + x^{2})\frac{dy}{dx}+2xy = f(x)\) can be rewritten as \(\frac{dy}{dx}+\frac{2x}{1 + x^{2}}y=\frac{f(x)}{1 + x^{2}}\). The integrating factor is \(e^{\int\frac{2x}{1 + x^{2}}dx}\). Let \(u = 1 + x^{2}\), then \(du=2xdx\), and \(\int\frac{2x}{1 + x^{2}}dx=\ln(1 + x^{2})\), so the integrating factor is \(1 + x^{2}\).

Step2: Solve for \(y\) on \(0\leq x<1\)

The general solution of the ODE is \(y(1 + x^{2})=\int f(x)dx+C\). For \(0\leq x<1\), \(f(x)=x\), so \(y(1 + x^{2})=\int xdx+C=\frac{x^{2}}{2}+C\). Using the initial condition \(y(0) = 0\), when \(x = 0,y = 0\), we get \(0=\frac{0^{2}}{2}+C\), so \(C = 0\). Then \(y=\frac{x^{2}}{2(1 + x^{2})}\) for \(0\leq x<1\).

Step3: Solve for \(y\) on \(x\geq1\)

We first need to find the value of \(y\) at \(x = 1\). When \(x = 1\), \(y=\frac{1^{2}}{2(1 + 1^{2})}=\frac{1}{4}\). For \(x\geq1\), \(f(x)=-x\), and the ODE gives \(y(1 + x^{2})=\int - xdx+C=-\frac{x^{2}}{2}+C\). Substitute \(x = 1,y=\frac{1}{4}\) into \(y(1 + x^{2})=-\frac{x^{2}}{2}+C\), we have \(\frac{1}{4}(1 + 1)=-\frac{1}{2}+C\), which gives \(C = 1\). So \(y=\frac{1-\frac{x^{2}}{2}}{1 + x^{2}}=\frac{2 - x^{2}}{2(1 + x^{2})}\) for \(x\geq1\).

Answer:

\(y=

$$\begin{cases}\frac{x^{2}}{2(1 + x^{2})},&0\leq x<1\\\frac{2 - x^{2}}{2(1 + x^{2})},&x\geq1\end{cases}$$

\)