Chapter 7.1, Problem 6E

To determine

To find the general solution to the exact differential equation.

Answer to Problem 6E

  $\text{The general solution is, }y={\sin }^{-1}x-2\sqrt{x}+C.$

Explanation of Solution

Given information:

  $\text{The given differential equation is, }\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{x}}.$

Calculation: The general solution solution is obtained as,

  $\begin{array}{l}\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{x}}\\ dy=\left(\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{x}}\right)dx\\ {\displaystyle \int dy={\displaystyle \int \left(\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{x}}\right)dx\left(\text{taking integral both side}\right)}}\\ \text{Then integrate each integral,}\\ y={\displaystyle \int \frac{1}{\sqrt{1-x^2}}dx-{\displaystyle \int x^{-\frac{1}{2}}dx}}\\ y={\sin }^{-1}x-2x^{\frac{1}{2}}+C\\ y={\sin }^{-1}x-2\sqrt{x}+C\\ \text{Hence, the general solution is, }y={\sin }^{-1}x-2\sqrt{x}+C.\end{array}$