Question

In Problems 31–34 verify that the indicated expression is an implicit

solution of the given differential equation.

${\left(\frac{dy}{dx}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{;}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{5}{\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{1}$

Short answer

The indicated expression is an implicit solution for the given differential equation.

Step-by-step solution

Definition of differential equation

A differential equation is defined as an equation that consists of the derivatives of one or more dependent functions with respect to one or more independent functions.

Taking derivative with respect to x

Given an implicit solution, $(x-5{)}^2+y^2=1$.

Differentiate the expression with respect to x on both sides and rewrite it to the given differential equation.

Taking the derivative with respect to x,

$2(x-5)+2y\frac{dy}{dx}=0$

Subtract both sides by $2(x-5)$and Divide both sides by 2, we get

$y\frac{dy}{dx}=-(x-5)$

Taking square on both sides,

$y^2{\left(\frac{dy}{dx}\right)}^2=(x-5{)}^2$

We know that, $(x-5{)}^2+y^2=1$.

So, $y^2{\left(\frac{dy}{dx}\right)}^2=1-y^2$.

Divide both sides by$y^2$,

$\begin{array}{c}{\left(\frac{dy}{dx}\right)}^2=\frac{1}{y^2}-1\\ {\left(\frac{dy}{dx}\right)}^2+1=\frac{1}{y^2}\end{array}$

Hence, verified.

Therefore, the indicated expression is an implicit solution for the given differential equation.