# First order separable Equation ODE

## 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 dont understand why

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Mark44
Mentor

## 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.
sanhuy said:
$$\left|5-y\right|\:=\:De^{-sinx}\:where\:D\:=\:e^{-C}$$
$$\left|5-y\right|\:=\:5-y\:,\:if\:5-y\:>\:0\:and\:\left|5-y\right|\:=\:-\left(5-y\right)\:if\:5-y<0$$
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
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
Mentor

## 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
$$y=5\:-De^{-sinx}\:\:and\:\:y\:=\:5+De^{-sinx}$$[/B]
$$y=5\:-2e^{-sinx}\:\:and\:\:y\:=\:5+2e^{-sinx}$$

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.

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
Mentor
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.

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
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.)

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
Mentor

haruspex
Homework Helper
Gold Member
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.

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