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

[physics604]

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.

 

[Dick]

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?

 

[physics604]

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

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

 

[Dick]

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?

 

[physics604]

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 $$y=\frac{e-e^y}{x}=\frac{e-e^y}{0}=undefined$$

 

[Dick]

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?

 

[physics604]

y=1. Okay, thanks!

 

[HallsofIvy]

First note thFirst note that when x= 0, $xy+ e^y= e^y= e$ so y(0)= 1. From $xy+ e^y= e$, yes, $xy'+ y+ e^yy'= 0$. When x= 0, $xy'+ y+ e^yy'= 0$ becomes $y+ e^yy'= 1+ ey'= 0$ so $y'(0)= -e^{-1}$.

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