Question

(a) Find an implicit solution of the initial-value problem,

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{{\mathbf{y}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{xy}}\mathbf{,}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}\sqrt{\mathbf{2}}$

(b) Find an explicit solution of the problem in part (a) and give

the largest interval I over which the solution is dened. A

graphing utility may be helpful here.;

Short answer

We will solve the given differential equation using implicit and explicit solution method.

Step-by-step solution

substituting the given differential equation.

$\begin{array}{l}y=ux\Rightarrow y\text{'}=\frac{dy}{dx}=u\text{'}x+u\\ u\text{'}x+u=\frac{u^2{x}^2-x^2}{u^2{x}^2}\\ u\text{'}x+u=\frac{u^2-1}{u}\\ u\text{'}x=\frac{u^2-1}{u}-u\\ u\text{'}x=\frac{u^2-1-u}{u}\\ \frac{du}{dx}x=-\frac{1}{u}\\ \int udu=-\int \frac{dx}{x}\\ \frac{u^2}{2}=-\ln \left|x\right|+c\\ u^2=-2\ln \left|x\right|+c_1\\ \frac{y^2}{x^2}=-2\ln \left|x\right|+c_1\\ y^2=-2x^2\ln \left|x\right|+x^2{c}_1\end{array}$

use the initial condition to solve

$\begin{array}{l}y\left(1\right)=-\sqrt{2}\\ (-\sqrt{2}{)}^2=-2\ln \left|1\right|+c_1\\ 2=c_1\end{array}$

substitute the value $2=c_1$

$y^2=-2x^2\ln \left|x\right|+2x^2$

finding the final answer by further solution

The implicit solution of initial value problem as obtained

$\begin{array}{l}{y}^2=-2x^2\ln \left|x\right|+2x^2\\ y^2=2x^2\left(1-\ln \left|x\right|\right)\end{array}$

The initial condition is satisfied

$\begin{array}{l}1-\ln \left|x\right|\ge 0\\ \ln \left|x\right|\le 1\\ \left|x\right|\le e^1\\ -e\le x\le e\end{array}$