Question

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

f(x)=\begin{cases}x, & 0leq x < 1\\-x, & xgeq1end{cases}$

Explanation:

Step1: Rewrite the differential equation

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

Step2: Solve the differential equation for \(0\leq x\lt1\)

The general solution of the linear differential equation \(\frac{dy}{dx}+P(x)y = Q(x)\) is \(y\cdot(I.F.)=\int Q(x)\cdot(I.F.)dx + C\). For \(0\leq x\lt1\), \(Q(x)=\frac{x}{1 + x^{2}}\) and \(I.F. = 1 + x^{2}\), so \(y(1 + x^{2})=\int xdx=\frac{x^{2}}{2}+C\). Using the initial condition \(y(0) = 0\), when \(x = 0,y = 0\), we get \(0=\frac{0}{2}+C\), so \(C = 0\). Then \(y=\frac{x^{2}}{2(1 + x^{2})}\) for \(0\leq x\lt1\).

Step3: Solve the differential equation for \(x\geq1\)

For \(x\geq1\), \(Q(x)=\frac{-x}{1 + x^{2}}\) and \(I.F. = 1 + x^{2}\). So \(y(1 + x^{2})=\int - xdx=-\frac{x^{2}}{2}+C\). To find \(C\), we need to ensure continuity at \(x = 1\). When \(x = 1\) from the left - hand side, \(y=\frac{1}{2(1 + 1)}=\frac{1}{4}\). Substituting \(x = 1,y=\frac{1}{4}\) into \(y(1 + x^{2})=-\frac{x^{2}}{2}+C\), we have \(\frac{1}{4}(1 + 1)=-\frac{1}{2}+C\), \(\frac{1}{2}=-\frac{1}{2}+C\), so \(C = 1\). Then \(y=\frac{1-\frac{x^{2}}{2}}{1 + x^{2}}=\frac{2 - x^{2}}{2(1 + x^{2})}\) for \(x\geq1\).

Answer:

\(y=

$$\begin{cases}\frac{x^{2}}{2(1 + x^{2})},&0\leq x\lt1\\\frac{2 - x^{2}}{2(1 + x^{2})},&x\geq1\end{cases}$$

\)