Solving for y'' at x=0 in the Equation $xy+e^y=e$

  • Thread starter physics604
  • Start date
In summary: That is, e^yy''(0)= -e^{-1}- 1+ e^{-1}+ 2e^{-2}= e^{-1}- 2e^{-2}. Finally, y''(0)= -1+ 2e^{-1}.In summary, the value of y'' at the point where x=0 is -1+ 2e^{-1}.
  • #1
physics604
92
2

Homework Statement



$$xy+e^y=e$$ Find the value of y'' at the point where x=0.

Homework Equations



Chain rule

The Attempt at a Solution



I find y' first.

$$x\frac{dy}{dx}+y+e^y\frac{dy}{dx}=0$$ $$(x+e^y)\frac{dy}{dx}=-y$$ $$\frac{dy}{dx}=\frac{-y}{x+e^y}$$

Then I find y''.

$$\frac{(x+e^y)(-\frac{dy}{dx})-(-y)(1+e^y\frac{dy}{dx})}{(x+e^y)^2}$$ $$\frac{(x+e^y)(\frac{y}{x+e^y})-(-y-ye^y(\frac{-y}{x+e^y}))}{(x+e^y)^2}$$ $$\frac{y-(-y+\frac{y^2e^y}{x+e^y})}{(x+e^y)^2}$$ $$\frac{y+y-\frac{y^2e^y}{x+e^y}}{(x+e^y)^2}$$ $$(2y-\frac{y^2e^y}{x+e^y})(\frac{1}{(x+e^y)^2})$$ $$\frac{2y(x+e^y)-y^2e^y}{(x+e^y)^3}$$ $$\frac{2xy+2ye^y-y^2e^y}{(x+e^y)^3}$$

x=0 so plugging that in I get

$$\frac{ye^y(2-y)}{e^{3y}}=\frac{y(2-y)}{e^{2y}}$$

The answer is $$\frac{1}{e^2}$$ What did I do wrong?

Any help is much appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
physics604 said:

Homework Statement



$$xy+e^y=e$$ Find the value of y'' at the point where x=0.

Homework Equations



Chain rule

The Attempt at a Solution



I find y' first.

$$x\frac{dy}{dx}+y+e^y\frac{dy}{dx}=0$$ $$(x+e^y)\frac{dy}{dx}=-y$$ $$\frac{dy}{dx}=\frac{-y}{x+e^y}$$

Then I find y''.

$$\frac{(x+e^y)(-\frac{dy}{dx})-(-y)(1+e^y\frac{dy}{dx})}{(x+e^y)^2}$$ $$\frac{(x+e^y)(\frac{y}{x+e^y})-(-y-ye^y(\frac{-y}{x+e^y}))}{(x+e^y)^2}$$ $$\frac{y-(-y+\frac{y^2e^y}{x+e^y})}{(x+e^y)^2}$$ $$\frac{y+y-\frac{y^2e^y}{x+e^y}}{(x+e^y)^2}$$ $$(2y-\frac{y^2e^y}{x+e^y})(\frac{1}{(x+e^y)^2})$$ $$\frac{2y(x+e^y)-y^2e^y}{(x+e^y)^3}$$ $$\frac{2xy+2ye^y-y^2e^y}{(x+e^y)^3}$$

x=0 so plugging that in I get

$$\frac{ye^y(2-y)}{e^{3y}}=\frac{y(2-y)}{e^{2y}}$$

The answer is $$\frac{1}{e^2}$$ What did I do wrong?

Any help is much appreciated.

You aren't finished. Same issue as in the last problem. Did you try using your function definition to simplify that?
 
  • #3
I've tried plugging in $$e^y=e-xy$$ but I don't get anywhere.

$$\frac{y(2-y)}{e^{2y}}=\frac{y(2-y)}{(e-xy)^2}$$

x=0 so

$$\frac{y(2-y)}{e^2}$$
 
  • #4
physics604 said:
I've tried plugging in $$e^y=e-xy$$ but I don't get anywhere.

$$\frac{y(2-y)}{e^{2y}}=\frac{y(2-y)}{(e-xy)^2}$$

x=0 so

$$\frac{y(2-y)}{e^2}$$

Put x=0 into your original equation. Can you conclude anything about the value of y when x=0?
 
  • #5
What do you mean?

If I plug x=0 into my original equation I get $$e^y=e-(0)y$$ so $$e^y=e$$ Or I can rearrange the equation to get $$y=\frac{e-e^y}{x}=\frac{e-e^y}{0}=undefined$$
 
  • #6
physics604 said:
What do you mean?

If I plug x=0 into my original equation I get $$e^y=e-(0)y$$ so $$e^y=e$$ Or I can rearrange the equation to get $$y=\frac{e-e^y}{x}=\frac{e-e^y}{0}=undefined$$

You should be able to figure out what y is when x=0 from e^y=e. What is it?
 
  • #7
y=1. Okay, thanks!
 
  • #8
First note thFirst note that when x= 0, [itex]xy+ e^y= e^y= e[/itex] so y(0)= 1. From [itex]xy+ e^y= e[/itex], yes, [itex]xy'+ y+ e^yy'= 0[/itex]. When x= 0, [itex]xy'+ y+ e^yy'= 0[/itex] becomes [itex]y+ e^yy'= 1+ ey'= 0[/itex] so [itex]y'(0)= -e^{-1}[/itex].

Differentiate again: [itex]xy''+ 2y'+ e^yy'+ e^yy''= 0[/itex]. When x= 0, y(0)= 1 and [itex]y'(0)= -e^{-1}[/itex] so [itex]-2e^{-1}- 1+ ey''(0)= 0[/itex]. [itex]ey'(0)= 1+ 2e^{-1}[/itex] so [itex]y'(0)= e^{-1}+ 2e^{-2}[/itex].
 

What is the equation for solving for y at x=0?

The equation for solving for y at x=0 is xy+e^y=e.

What does it mean to solve for y at x=0?

Solving for y at x=0 means finding the value of y when x=0 in the given equation.

How do I solve for y at x=0 in this equation?

To solve for y at x=0 in the equation xy+e^y=e, you can use algebraic manipulation or use a graphing calculator to find the intersection point of the equation with the x-axis.

Why is solving for y at x=0 important?

Solving for y at x=0 is important because it helps find a specific solution to the equation at a specific point, which can provide insights and understanding into the behavior of the equation.

What are the applications of solving for y at x=0 in this equation?

Solving for y at x=0 in this equation can be used in various fields such as physics, engineering, and economics to understand and predict behaviors and outcomes of systems represented by the equation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
558
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
478
  • Calculus and Beyond Homework Help
Replies
6
Views
793
  • Calculus and Beyond Homework Help
Replies
6
Views
695
  • Calculus and Beyond Homework Help
Replies
4
Views
635
  • Calculus and Beyond Homework Help
Replies
5
Views
705
  • Calculus and Beyond Homework Help
Replies
7
Views
641
  • Calculus and Beyond Homework Help
Replies
8
Views
708
  • Calculus and Beyond Homework Help
Replies
2
Views
401
Back
Top