Solve the equation y' = x/r? + 1/(ye") and graph sev- eral members of the family of solutions (if your CAS does implicit plots). How does the solution curve change as the constant C varies?

the given differential equation is:

$y\text{'}=\frac{x\sqrt{x^2+1}}{y{e}^y}$

we have to graph the several members of the family of solutions.

then we have to determine how does the solution curve change as the constant C varies.

the given differential equation can be written as:

$\begin{array}{rcl}y\text{'}& =& \frac{x\sqrt{x^2+1}}{y{e}^y}\\ \frac{dy}{dx}& =& \frac{x\sqrt{x^2+1}}{y{e}^y}\\ y{e}^{y}dy& =& x\sqrt{x^2+1}dx \left(1\right)\end{array}$

now integrate the both sides of the equation (1),

$\int y{e}^{y}dy=\int x\sqrt{x^2+1}dx \left(2\right)$

let the integral $\int y{e}^{y}dy$ be $I_1$

therefore,

$I_1=\int y{e}^{y}dy$

integrate the integral $I_1$ by using the integration by parts.

therefore,

$\begin{array}{rcl}{I}_1& =& \int y{e}^{y}dy\\ & =& y\int e^{y}dy-\int \left(\frac{dy}{dy}\int e^{y}dy\right)dy\\ & =& y\left(e^y\right)-\int 1\times e^{y}dy\\ & =& y{e}^y-e^y\end{array}$

now let the integral $\int x\sqrt{x^2+1}dx$ be $I_2$

therefore,

$I_2=\int x\sqrt{x^2+1}dx$

let $x^2+1=t$

therefore,

$\begin{array}{rcl}d\left(x^2+1\right)& =& dt\\ 2xdx& =& dt\\ xdx& =& \frac{dt}{2}\end{array}$

now substitute these values in the integral $I_2$

therefore,

$\begin{array}{rcl}{I}_2& =& \int x\sqrt{x^2+1}dx\\ & =& \int \sqrt{t}\frac{dt}{2}\\ & =& \frac{1}{2}\int t^{\frac{1}{2}}dt\\ & =& \frac{1}{2}\left[\frac{t^{{\displaystyle \frac{1}{2}+1}}}{{\displaystyle \frac{1}{2}+1}}\right]\\ & =& \frac{1}{2}\times \frac{t^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\\ & =& \frac{1}{2}\times \frac{2}{3}{t}^{\frac{3}{2}}\\ & =& \frac{1}{3}{t}^{\frac{3}{2}}\end{array}$