# First order separable Equation ODE

Tags:
1. Jan 11, 2017

### sanhuy

1. The problem statement, all variables and given/known data
$$\frac{dy}{dx}\:+\:ycosx\:=\:5cosx$$
I get two solutions for y however only one of them is correct according to my online homework
(see attempt at solution)

2. Relevant equations
y(0) = 7 is initial condition

3. The attempt at a solution
$$\int \:\frac{1}{5-y}dy\:=\:\int \:cosxdx\:+\:C$$
$$-ln\left|5-7\right|\:=\:sinx\:+C$$
$$\left|5-y\right|\:=\:De^{-sinx}\:where\:D\:=\:e^{-C}$$
$$\left|5-y\right|\:=\:5-y\:,\:if\:5-y\:>\:0\:and\:\left|5-y\right|\:=\:-\left(5-y\right)\:if\:5-y<0$$
using initial conditions for both equations
$$y=5\:-De^{-sinx}\:\:and\:\:y\:=\:5+De^{-sinx}$$
$$y=5\:-2e^{-sinx}\:\:and\:\:y\:=\:5+2e^{-sinx}$$

however the first solution is incorrect I dont understand why

Last edited by a moderator: Jan 11, 2017
2. Jan 11, 2017

### Staff: Mentor

Did you mean 5 - y above? If you really meant 5 -7, I wouldn't substitute in the initial condition yet.
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. Jan 11, 2017

### Staff: Mentor

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. Jan 11, 2017

### sanhuy

Since 2e^−sinx>0 you can drop the absolute sign correct?
then it becomes
y=5−2e^−sinx?

5. Jan 11, 2017

### Staff: Mentor

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. Jan 11, 2017

### sanhuy

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. Jan 11, 2017

### Staff: Mentor

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. Jan 11, 2017

### sanhuy

both solutions work, but my webwork says the first one is wrong

9. Jan 11, 2017

### Staff: Mentor

10. Jan 11, 2017

### haruspex

They satisfy the differential equation, but I do not see how your negative multiplier solution satisfies the initial condition.

11. Jan 11, 2017

### sanhuy

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. Jan 11, 2017

### haruspex

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.