Separation of Variables : y'=sin(x)cos(y)

[VincentNY]

Hello,

I'm trying to find the solution by separation of variables for y'=sin(x)cos(y). The book (Kaplan Advanced Calculus) gives the answer: (1+sin y)=c cos y e^{-cos x} . I can't seem to figure out the steps to get this answer.

First I end up with dy/cos y=sin x dx.

I tried multiplying both sides by -sin y to get:

-siny dy/ cos y = -sin y sin x dx

u=cos y
du=-sinydy


the left side can now be integrated: ln |cos y| + C, I'm not sure how to handle the other side.

Any suggestions?

Thanks

V

 

[HallsofIvy]

Hello,

I'm trying to find the solution by separation of variables for y'=sin(x)cos(y). The book (Kaplan Advanced Calculus) gives the answer: (1+sin y)=c cos y e^{-cos x} . I can't seem to figure out the steps to get this answer.

First I end up with dy/cos y=sin x dx.

I tried multiplying both sides by -sin y to get:

-siny dy/ cos y = -sin y sin x dx

So having separated variables you immediately unseparate them? that makes no sense!

u=cos y
du=-sinydy


the left side can now be integrated: ln |cos y| + C, I'm not sure how to handle the other side.

Any suggestions?

Thanks

V

Of course, you cannot integrate the right side because you have both "x" and "y" and one is a function of the other! Instead treat the left side on its own:
\(\displaystyle \int \frac{dy}{cos(y)}= \int \frac{cos(y)dy}{cos^2(y)}= \int \frac{cos(y)dy}{1- sin^2(y)}\). Now use the substitution u= sin(y).

 

[soroban]

Hello, VincentNY!

\(\displaystyle \text{Solve: }\:\dfrac{dy}{dx} \:=\:\sin(x)\cos(y)\).

\(\displaystyle \text{Book answer: }\:1+\sin y \:=\:C\,\cos y\:e^{-\cos x}\)

Separate variables: .\(\displaystyle \dfrac{dy}{\cos y} \:=\:\sin x\,dx \quad\Rightarrow\quad \sec y\,dy \:=\:\sin x\,dx\)

Integrate: .\(\displaystyle \displaystyle \int \sec y\,dy \:=\:\int \sin x\,dx\)

. . . \(\displaystyle \ln|\sec y + \tan y| \;=\;-\cos x + c\)

. . .\(\displaystyle \ln\left|\dfrac{1}{\cos y} + \dfrac{\sin y}{\cos y}\right| \;=\;-\cos x + c\)

. . . . . . \(\displaystyle \ln\left|\dfrac{1+\sin y}{\cos y}\right| \;=\;-\cos x + c\)

. . . . . . . . \(\displaystyle \dfrac{1+\sin y}{\cos y} \;=\;e^{-\cos x \:+\: c} \;=\;e^{-\cos x}\cdot e^c\)

. . . . . . . . \(\displaystyle \dfrac{1+\sin y}{\cos y} \;=\;C\,e^{-\cos x}\)

. . . . . . . . \(\displaystyle 1 + \sin y \;=\;C\,\cos y\:e^{-\cos x}\)

 

[VincentNY]

So having separated variables you immediately unseparate them? that makes no sense!
Of course, you cannot integrate the right side because you have both "x" and "y" and one is a function of the other! Instead treat the left side on its own:
\(\displaystyle \int \frac{dy}{cos(y)}= \int \frac{cos(y)dy}{cos^2(y)}= \int \frac{cos(y)dy}{1- sin^2(y)}\). Now use the substitution u= sin(y).

HallsofIvy,

Thanks for your suggestion... I'm a little rusty. I have observed Soroban's solution below but decided to also try it with the substitution for kicks.

Let u= siny
let du =cosydy


From an integral table, I used:
let a=1
int [du/(a²-u²)=(1/2a)ln|(u+a)/(u-a)| + C

cosy/(1-sin²y)=sinxdx
int [cosy/(1²-siny²)=int[sinxdx]

1/2*ln|(siny+1)/(siny-1)|=-cosx+C

ln|(siny+1)/(siny-1)|=-2cosx+C

(siny+1)/(siny-1)=C^-2cosx

siny+1=C(siny-1)^-2cosx

 

[VincentNY]

Hi Soroban,

Thanks for your response!

 

[HallsofIvy]

HallsofIvy,

Thanks for your suggestion... I'm a little rusty. I have observed Soroban's solution below but decided to also try it with the substitution for kicks.

Let u= siny
let du =cosydy


From an integral table, I used:
let a=1
int [du/(a²-u²)=(1/2a)ln|(u+a)/(u-a)| + C

cosy/(1-sin²y)=sinxdx
int [cosy/(1²-siny²)=int[sinxdx]

1/2*ln|(siny+1)/(siny-1)|=-cosx+C

ln|(siny+1)/(siny-1)|=-2cosx+C

(siny+1)/(siny-1)=C^-2cosx

You are doin this wrong. Taking the exponential of both sides (inverse ln) gives \(\displaystyle e^{-2cosx+ C}= e^{-2cosx}e^C= ce^{-2cosx}\) where "\(\displaystyle c= e^C\)"

siny+1=C(siny-1)^-2cosx

And now sin y+ 1= ce^{-2cosx}(sin y- 1)[/tex], NOT (sin y- 1) to a power.