Chapter 6.3, Problem 15E

To determine

To find: The initial equation.

Answer to Problem 15E

The initial equation: $y=\frac{2}{3}x{(x-1)}^{3/2}-\frac{4}{15}{(x-1)}^{5/2}+2$

Explanation of Solution

Given information:

The differential equation is: $\frac{dy}{dx}=x\sqrt{x-1}$

And $y=2$ when

  $x=1$

Calculation:

The given differential equation is: $\frac{dy}{dx}=x\sqrt{x-1}$

Integrate each side after separating the variables. Use part-by-part integration

  $\begin{array}{}$ \begin{array}{c}\frac{dy}{dx}=x\sqrt{x-1}\\ dy=x\sqrt{x-1}dx\\ \int dy=\int x\sqrt{x-1}dx\end{array}$$ y=x\cdot \frac{2}{3}{(x-1)}^{3/2}-\int \frac{2}{3}{(x-1)}^{3/2}dx$$ y=\frac{2}{3}x{(x-1)}^{3/2}-\frac{2}{3}\int {(x-1)}^{3/2}dx$$ =\frac{2}{3}x{(x-1)}^{3/2}-\frac{2}{3}\left(\frac{{(x-1)}^{5/2}}{5/2}\right)+C$$ =\frac{2}{3}x{(x-1)}^{3/2}-\frac{2}{3}\left(\frac{2}{5}{(x-1)}^{5/2}\right)+C$$ =\frac{2}{3}x{(x-1)}^{3/2}-\frac{4}{15}{(x-1)}^{5/2}+C$\end{array}$

To find $C$ , use the initial conditions.

And $y=2$ when

  $x=1$

  $\begin{array}{c}$ y=\frac{2}{3}x{(x-1)}^{3/2}-\frac{4}{15}{(x-1)}^{5/2}+C$$ 2=\frac{2}{3}(1){(1-1)}^{3/2}-\frac{4}{15}{(1-1)}^{5/2}+C$$ 2=\frac{2}{3}{(0)}^{3/2}-\frac{4}{15}{(0)}^{5/2}+C$2=C\end{array}$

The equation is then $y=\frac{2}{3}x{(x-1)}^{3/2}-\frac{4}{15}{(x-1)}^{5/2}+2$

The slope field and the function's graph do match.

  

Therefore the required initial equation of given differential equation $\frac{dy}{dx}=x\sqrt{x-1}$ is

  $y=\frac{2}{3}x{(x-1)}^{3/2}-\frac{4}{15}{(x-1)}^{5/2}+2$ .