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

the given differential equation is: y'=xx2+1yey we have to graph the several members of the family…

Answered: Solve the equation y' = x/r? + 1/(ye")…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Related questions

Question

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?

Expert Solution

Step 1

the given differential equation is:

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

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:

$\begin{array}{rcl} y' & = & \frac{x \sqrt{x^{2} + 1}}{y e^{y}} \\ \frac{d y}{d x} & = & \frac{x \sqrt{x^{2} + 1}}{y e^{y}} \\ y e^{y} d y & = & x \sqrt{x^{2} + 1} d x (1) \end{array}$

Step 2

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

$\int y e^{y} d y = \int x \sqrt{x^{2} + 1} d x (2)$

let the integral $\int y e^{y} d y$ be $I_{1}$

therefore,

$I_{1} = \int y e^{y} d y$

integrate the integral $I_{1}$ by using the integration by parts.

therefore,

$\begin{array}{rcl} I_{1} & = & \int y e^{y} d y \\ = & y \int e^{y} d y - \int (\frac{d y}{d y} \int e^{y} d y) d y \\ = & y (e^{y}) - \int 1 \times e^{y} d y \\ = & y e^{y} - e^{y} \end{array}$

Step 3

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

therefore,

$I_{2} = \int x \sqrt{x^{2} + 1} d x$

let $x^{2} + 1 = t$

therefore,

$\begin{array}{rcl} d (x^{2} + 1) & = & d t \\ 2 x d x & = & d t \\ x d x & = & \frac{d t}{2} \end{array}$

now substitute these values in the integral $I_{2}$

Step 4

therefore,

$\begin{array}{rcl} I_{2} & = & \int x \sqrt{x^{2} + 1} d x \\ = & \int \sqrt{t} \frac{d t}{2} \\ = & \frac{1}{2} \int t^{\frac{1}{2}} d t \\ = & \frac{1}{2} [\frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}] \\ = & \frac{1}{2} \times \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \\ = & \frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}} \\ = & \frac{1}{3} t^{\frac{3}{2}} \end{array}$