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11.2 Initial value problem in explicit form.

karush

Well-known member
Jan 31, 2012
2,886
11.2.PNG

Given #11
$\quad\displaystyle
xdx+ye^{-x}dy=0,\quad y(0)=1$
a. Initial value problem in explicit form.
$\quad xdx=-ye^{-x}dy$
separate
$\quad \frac{x}{e^{-x}}\, dx=-y\, dy$
simplify
$\quad xe^x\, dx=-y\, dy$
rewrite
$\quad y\,dy=-xe^x\,dx$
integrate (with boundaries)
$\quad \int_1^y u\,dy=-\int_0^x ve^v\,dv$

OK i did this so far hopefully ok but didnt know how to do b and c (on desmos)

b. Plot the graph of the solution
c. Interval of solution.
 
Last edited:

Cbarker1

Active member
Jan 8, 2013
255
Re: 11.2

I recommend that you integrate both side to find the curve and the graph as well as provide that answer to us.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Re: 11.2

(You have a typo- your "dy" should be "du".)

Why stop there? Obviously you have to do the integrals! The "du" integral is straight forward. To do the "dv" integral, use "integration by parts".

Solving for y will involve a square root. The domain is determined by the fact that the square root must be of a non-negative number. The graph will have two parts, one with "+", the other with "-".
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: 11.2

What title would you like me to give this thread so that the title is useful?
 

karush

Well-known member
Jan 31, 2012
2,886
Re: 11.2

11.2 Initial value problem in explicit form.

I tried to fix it but couldn't

==================================

integrate (with boundaries)
$$\int_1^y y\,dy=-\int_0^x xe^x\,dx$$
then
$\frac{y^2}{2}-\frac{1}{2}=-e^xx+e^x-1$
isolate y
multiply thru by 2 and add 1
$y^2=-2e^xx+2e^x-2+1$
simplify RHS and factor
$y^2=2[1-x]e^x-1$
square both sides
$y = 2[(1-x)e^x-1]^{1/2}$

typo maybe??

when this is graphed you get nothing????
 
Last edited:

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,180
Re: 11.2


thake the square (root) of both sides
$y = 2[(1-x)e^x-1]^{1/2}$

typo maybe??

when this is graphed you get nothing????
The "2" belongs under the square root. Typo? (Sun)

Here's the graph. I have no idea why this would cause you trouble.
graph.jpg

And don't forget the part of the solution with the negative sign!

-Dan
 

karush

Well-known member
Jan 31, 2012
2,886
11.2d.PNG
here is the desmos attempt


also I don't know where the interval of $−1.68<x<0.77$ which is the book answer comes from
I set y=0 but that didn't return those values
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
here is the desmos attempt


also I don't know where the interval of $−1.68<x<0.77$ which is the book answer comes from
I set y=0 but that didn't return those values
What values did you get? If I graph the function on Desmos (https://www.desmos.com/calculator) clicking on the y-intercept, I get x= -1.678 and x= 0.768)

If y= 0 then [tex]2(1- x)e^x- 1= 0[/tex] then [tex]2(1- x)e^x= 1[/tex].

Let u= 1- x so that x= 1- u. The equation becomes [tex]2ue^{1- u}= 2eue^{-u}= 1[/tex]. Let v= -u so that u= -v. The equation becomes [tex]-2eve^v= 1[/tex]. Divide both sides by -2e. [tex]ve^v= -\frac{1}{2e}[/tex]. Then [tex]v= W\left(\frac{-1}{2e}\right)[/tex] where "W" is "Lambert's W function, defined as the inverse function to [tex]f(x)= xe^x[/tex].