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First order separable Equation ODE

  • #1
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2

Homework Statement


[tex] \frac{dy}{dx}\:+\:ycosx\:=\:5cosx [/tex]
I get two solutions for y however only one of them is correct according to my online homework
(see attempt at solution)

Homework Equations


y(0) = 7 is initial condition

The Attempt at a Solution


[tex]\int \:\frac{1}{5-y}dy\:=\:\int \:cosxdx\:+\:C [/tex]
[tex] -ln\left|5-7\right|\:=\:sinx\:+C [/tex]
[tex] \left|5-y\right|\:=\:De^{-sinx}\:where\:D\:=\:e^{-C} [/tex]
[tex] \left|5-y\right|\:=\:5-y\:,\:if\:5-y\:>\:0\:and\:\left|5-y\right|\:=\:-\left(5-y\right)\:if\:5-y<0 [/tex]
using initial conditions for both equations
[tex] y=5\:-De^{-sinx}\:\:and\:\:y\:=\:5+De^{-sinx}[/tex]
[tex] y=5\:-2e^{-sinx}\:\:and\:\:y\:=\:5+2e^{-sinx} [/tex]

however the first solution is incorrect I dont understand why
 
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Answers and Replies

  • #2
33,084
4,787

Homework Statement


[tex] \frac{dy}{dx}\:+\:ycosx\:=\:5cosx [/tex]
I get two solutions for y however only one of them is correct according to my online homework
(see attempt at solution)

Homework Equations


y(0) = 7 is initial condition

The Attempt at a Solution


[tex]\int \:\frac{1}{5-y}dy\:=\:\int \:cosxdx\:+\:C [/tex]
[tex] -ln\left|5-7\right|\:=\:sinx\:+C [/tex]
Did you mean 5 - y above? If you really meant 5 -7, I wouldn't substitute in the initial condition yet.
sanhuy said:
[tex] \left|5-y\right|\:=\:De^{-sinx}\:where\:D\:=\:e^{-C} [/tex]
[tex] \left|5-y\right|\:=\:5-y\:,\:if\:5-y\:>\:0\:and\:\left|5-y\right|\:=\:-\left(5-y\right)\:if\:5-y<0 [/tex]
using initial conditions for both equations
[tex] y=5\:-De^{-sinx}\:\:and\:\:y\:=\:5+De^{-sinx}[/tex]
[tex] y=5\:-2e^{-sinx}\:\:and\:\:y\:=\:5+2e^{-sinx} [/tex]

however the first solution is incorrect I dont understand why
I'm not totally clear on what you're doing above. Since y(0) = 7, you can assume that y -7 < 0 for solutions near the initial condition.
 
  • #3
12,338
8,727

Homework Statement


[tex] \frac{dy}{dx}\:+\:ycosx\:=\:5cosx [/tex]
I get two solutions for y however only one of them is correct according to my online homework
(see attempt at solution)

Homework Equations


y(0) = 7 is initial condition

The Attempt at a Solution


[tex]\int \:\frac{1}{5-y}dy\:=\:\int \:cosxdx\:+\:C [/tex]
[tex] -ln\left|5-7\right|\:=\:sinx\:+C [/tex]
[tex] \left|5-y\right|\:=\:De^{-sinx}\:where\:D\:=\:e^{-C} [/tex]
[tex] \left|5-y\right|\:=\:5-y\:,\:if\:5-y\:>\:0\:and\:\left|5-y\right|\:=\:-\left(5-y\right)\:if\:5-y<0 [/tex]
using initial conditions for both equations
[tex] y=5\:-De^{-sinx}\:\:and\:\:y\:=\:5+De^{-sinx}[/tex][/B]
[tex] y=5\:-2e^{-sinx}\:\:and\:\:y\:=\:5+2e^{-sinx} [/tex]

however the first solution is incorrect I dont understand why
Why don't you apply the initial condition on ##\left|5-y\right|\:=\:De^{-sinx}\,##? That gives ##\left|5-y\right|\:=\:2e^{-sinx} > 0##. Now check your signs again.
 
  • #4
40
2
Why don't you apply the initial condition on ##\left|5-y\right|\:=\:De^{-sinx}\,##? That gives ##\left|5-y\right|\:=\:2e^{-sinx} > 0##. Now check your signs again.
Since 2e^−sinx>0 you can drop the absolute sign correct?
then it becomes
y=5−2e^−sinx?
 
  • #5
12,338
8,727
Since 2e^−sinx>0 you can drop the absolute sign correct?
then it becomes
y=5−2e^−sinx?
No you can't, because inside the absolute sign can still happen both possibilities. Just consider the cases ##y(x) > 5## and ##y(x) \leq 5##.

Edit: Sorry, I think I made a sign error myself.
 
  • #6
40
2
No you can't, because inside the absolute sign can still happen both possibilities. Just consider the cases ##y(x) > 5## and ##y(x) \leq 5##.
if y(x) > 5 then |y-5| = (y-5)
and if y(x) <= 5 we get |y-5| = -(y-5) ?
then we get two solutions that i posted above again? Am i missing something here
 
  • #7
12,338
8,727
Now, that I've seen my sign error, I did what you should do as well with all of these kind of equations: check them by differentiation. (I think you were right in the first place.)
 
  • #8
40
2
Now, that I've seen my sign error, I did what you should do as well with all of these kind of equations: check them by differentiation. (I think you were right in the first place.)
both solutions work, but my webwork says the first one is wrong
 
  • #9
12,338
8,727
Maybe you've had a typo in your other solution. Or some additional information is hidden somewhere in your text.
 
  • #10
haruspex
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both solutions work, but my webwork says the first one is wrong
They satisfy the differential equation, but I do not see how your negative multiplier solution satisfies the initial condition.
 
  • #11
40
2
They satisfy the differential equation, but I do not see how your negative multiplier solution satisfies the initial condition.
I just noticed this as well. BTW I was just wondering the way I got rid of my absolute values was correct right, regardless of the general problem i stated?
 
  • #12
haruspex
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I just noticed this as well. BTW I was just wondering the way I got rid of my absolute values was correct right, regardless of the general problem i stated?
I was not quite sure what you did. I would substitute x=0, y=7 with still |y-5| to find D=2. Then you get y=5+/-2e-sin(x), then reuse the initial condition to rule out the -2 case.
 

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