Question

i. solve the differential equation $\frac{dy}{dx}=x\sqrt{1 + x^{2}}$ if $y = - 2$ when $x = 0$.

Explanation:

Step1: Separate variables

We have \(\frac{dy}{dx}=x\sqrt{1 + x^{2}}\), which can be rewritten as \(dy=x\sqrt{1 + x^{2}}dx\).

Step2: Integrate both sides

Integrate the right - hand side using substitution. Let \(u = 1+x^{2}\), then \(du=2xdx\) and \(xdx=\frac{1}{2}du\).
The integral \(\int x\sqrt{1 + x^{2}}dx=\frac{1}{2}\int\sqrt{u}du\).
Using the power rule \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n=\frac{1}{2}\)), we get \(\frac{1}{2}\times\frac{2}{3}u^{\frac{3}{2}}+C=\frac{1}{3}(1 + x^{2})^{\frac{3}{2}}+C\).
Integrating the left - hand side \(\int dy=y\). So \(y=\frac{1}{3}(1 + x^{2})^{\frac{3}{2}}+C\).

Step3: Use the initial condition

Given \(y=-2\) when \(x = 0\). Substitute \(x = 0\) and \(y=-2\) into \(y=\frac{1}{3}(1 + x^{2})^{\frac{3}{2}}+C\).
We have \(-2=\frac{1}{3}(1+0)^{\frac{3}{2}}+C\), which simplifies to \(-2=\frac{1}{3}+C\). Solving for \(C\), we get \(C=-2-\frac{1}{3}=-\frac{7}{3}\).

Answer:

\(y=\frac{1}{3}(1 + x^{2})^{\frac{3}{2}}-\frac{7}{3}\)