First order separable Equation ODE

[sanhuy]

Homework Statement

\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)

Homework Equations

y(0) = 7 is initial condition

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 don't understand why

 

[Mark44]

Homework Statement

\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)

Homework Equations

y(0) = 7 is initial condition

The Attempt at a Solution

\int \:\frac{1}{5-y}dy\:=\:\int \:cosxdx\:+\:C
-ln\left|5-7\right|\:=\:sinx\:+C

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

\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 don't 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.

 

[fresh_42]

Homework Statement

\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)

Homework Equations

y(0) = 7 is initial condition

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\:&gt;\:0\:and\:\left|5-y\right|\:=\:-\left(5-y\right)\:if\:5-y&lt;0
using initial conditions for both equations
y=5\:-De^{-sinx}\:\:and\:\:y\:=\:5+De^{-sinx}[/B]
y=5\:-2e^{-sinx}\:\:and\:\:y\:=\:5+2e^{-sinx}

however the first solution is incorrect I don't 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.

 

[sanhuy]

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?

 

[fresh_42]

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.

 

[sanhuy]

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

 

[fresh_42]

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.)

 

[sanhuy]

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

 

[fresh_42]

Maybe you've had a typo in your other solution. Or some additional information is hidden somewhere in your text.

 

[haruspex]

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.

 

[sanhuy]

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?

 

[haruspex]

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.