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