Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I solve the differential equation: $$y'=\sin(x-y)$$

Could I do this?

$$\frac{dy}{dx}= \sin x \cos y - \sin y \cos x$$

But, how would I continue?

share|improve this question
    
Wolfram Alpha provides the answers obtained in: wolframalpha.com/input/?i=y%27+%3D+sin%28x-y%29 –  Leucippus Jul 14 at 22:22
    
@Leucippus I found that the general solution is: $$\tan{(x-y)}+ \sec{(x-y)}=x+c$$ Is this equal to the one that wikipedia provides? –  evinda Jul 14 at 23:04
1  
What is provided from the Wolfram solution is the complete solution of $y(x)$. The solution you have provided, which works, is a mixed solution containing a function of $x$ and $y(x)$. Solving the solution you have provided for $y(x)$ should provide something close to that provided by Wolfram. –  Leucippus Jul 14 at 23:36
    
@Leucippus But how can I solve it for $y(x)$ ? –  evinda Jul 14 at 23:53

3 Answers 3

If you substitute $y_1=y-x$ you get:

$$y_1'=y'-1$$

$$\sin(x-y)=\sin(x-y_1-x)=\sin(y_1)$$

So:

$$\sin(y_1)=y_1'+1$$

Next you get $1=\dfrac{y_1'}{\sin(y_1)-1}$ and you can integrate both sides.

You get:

$1)$ Left side: $$\int_{x_0}^{x} 1 dx=x-x_0$$

$2)$Right side (by changing variables): $$\int_{x_0}^{x} \frac{y_1'(x)}{\sin(y_1(x))-1} dx=\int_{y_1(x_0)}^{y(x)} \frac{1}{\sin x-1} dx$$

share|improve this answer
1  
Isn't it $\sin(-y_1)=-\sin(y_1)$ ? So,do we have to substitute maybe : $y_1=y-x$ ? –  evinda Jul 14 at 21:49
    
If $y_1 = y + x$ then $x - y = 2x - y_1$. You should use $y_1 = x-y$... –  johannesvalks Jul 14 at 21:51
    
So,we do it like that: $$u=x-y$$ $$u'=1-y'$$ $$1-u'= \sin u \Rightarrow 1- \sin u=u' \Rightarrow \frac{du}{dx}=1- \sin u \Rightarrow \frac{du}{ 1- \sin u}=dx$$ But how can I continue? –  evinda Jul 14 at 21:53
1  
$\int\frac{1}{1-\sin(u)}du=\int\frac{1+\sin(u)}{\cos^{2}(u)}du=\int\sec^{2}(u)du‌​+\int\tan(u)\sec(u)du$ –  user71352 Jul 14 at 21:57
1  
@mesel Agreed. (+1 from me, in any event; I wasn't a downvoter, though) –  anorton Jul 15 at 0:35

I will only add the solution of the integral,

$$\int\dfrac{1}{1-\sin x}dx$$ set $\tan(x/2)=u$ then you will have

$$\int\dfrac{1}{1-{2u\over{1+u^2}}} \dfrac{2}{u^2+1} du$$

As a result you will have $$\dfrac{2\sin(\dfrac x2)}{\cos(\dfrac x2)-\sin(\dfrac x2)}+c$$

share|improve this answer
    
How do you get from the first integral to the seond one,using the substitution? Do you use a formula? –  evinda Jul 14 at 22:14
    
@evinda: just the substition, if $tan(x/2)=u$ then $cos(x/2)=\dfrac {1}{\sqrt{1+u^2}}$ and $sin(x/2)=\dfrac {u}{\sqrt{1+u^2}}$ then use $sin(x)=2sin(x/2)cos(x/2)=\dfrac {2u}{{1+u^2}}$ –  mesel Jul 14 at 22:17
    
A ok..is the general solution $\tan(x-y)+ \sec(x-y)=x+c$,or have I found it wrong? –  evinda Jul 14 at 22:40
1  
It's $\large{2\,{\rm d}u \over 1 + u^{2}}$ in your second line. –  Felix Marin Jul 14 at 23:32
    
@FelixMarin: thank you for your correction. –  mesel Jul 15 at 0:30

Let $z = x - y$. Thus, $\dfrac{dy}{dx} = 1 - \dfrac{dz}{dx}$ and $$ \dfrac{dy}{dx} = \sin(x - y) \quad \Rightarrow \quad \dfrac{dz}{dx} = 1 - \sin z \quad \Rightarrow \quad \int dx = \int \dfrac{dz}{1 -\sin z} $$ The next step is to change $u = \tan(z/2)$ so that $dz = \dfrac{2du}{1 + u^2}$. Note that $$ \sin z = \dfrac{2\sin(z/2)\cos(z/2)}{\cos^2(z/2) + \sin^2(z/2)} = \dfrac{2\tan(z/2)}{1 + \tan^2(z/2)} = \dfrac{2u}{1 + u^2} $$ Thus, $$ x = \int \dfrac{\dfrac{2du}{1 + u^2}}{1 -\dfrac{2u}{1 + u^2}} =2\int \dfrac{du}{1 + u^2 - 2u} = 2\int \dfrac{du}{(1-u)^2} = -\dfrac{2}{1-u} + C \quad \Rightarrow $$ $$ x = \dfrac{2}{\tan(z/2) - 1} + C = \dfrac{2}{\tan\biggl(\dfrac{x - y}{2}\biggr) - 1} + C $$

share|improve this answer
    
So,is the general solution $\tan(x-y)+ \sec(x-y)=x+c$ ? –  evinda Jul 14 at 22:40
    
Write $z = x - y$ and proceeds as before. –  MathFacts Jul 14 at 23:08
    
Could you explain it further to me? Can we not solve this equation for $y(x)$ ? –  evinda Jul 14 at 23:55
    
As before, I think the solution that other differential equation can not be explained to x. –  MathFacts Jul 14 at 23:58
    
At the book,it is given the differential equaltion $y'=\sin (x-y) \text{ and } y(0)=\pi$ and the solution is $y=x-2 \arctan(x+c) + \frac{\pi}{2}$..How can I conclude to this? Substituting $x=0$ at the solution that I found, I found that $c=-1$..How could I continue? –  evinda Jul 15 at 0:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.