Solving Differential Equation: y' = cos(x+y) with u = x+y

[der.physika]

Solve the following differential equation

y\prime=\cos(x+y)

Here introduce the new variable:

x+y\equiv{u}

Please show steps, or else I won't understand this

 

[LCKurtz]

What do you get when you differentiate x + y = u with respect to x and plug it in?

 

[der.physika]

What do you get when you differentiate x + y = u with respect to x and plug it in?

Okay, but how do I differentiate

<br /> x+y\equiv{u}<br /> ?

Do I get

<br /> 1+y\equiv{u}<br />

or do I have to rewrite it as

<br /> u-x\equiv{y}<br />

which then becomes

<br /> u-1\equiv{y\prime}<br />

and where do I plug that in?

 

[LCKurtz]

Okay, but how do I differentiate

<br /> x+y\equiv{u}<br /> ?

Do I get

<br /> 1+y\equiv{u}<br />

or do I have to rewrite it as

<br /> u-x\equiv{y}<br />

which then becomes

<br /> u-1\equiv{y\prime}<br />

You differentiate an equation by differentiating both sides of it. Every term. Neither of your two attempts to differentiate above are correct.

and where do I plug that in?

Look at your original post. Where do you think you might plug the substitution in?

 

[der.physika]

ok how about this

<br /> x+y\equiv{u}<br />

comes out to be

<br /> 1+y\prime\equiv{u\prime}<br />

and then plug the y\prime into the equation?

 

[LCKurtz]

Yes. Your new equation should involve u and x.

 

[der.physika]

Yes. Your new equation should involve u and x.

ok, so I plugged it in and I got the following

u\prime-1=\cos(u)

what do I do now? should I integrate both sides? and then rewrite u into y and x terms?

 

[LCKurtz]

ok, so I plugged it in and I got the following

u\prime-1=\cos(u)

what do I do now? should I integrate both sides? and then rewrite u into y and x terms?

Well, you have a new differential equation. What methods have you learned to solve differential equations? Have you tried any of these methods? Show us what happens.

 

[LCKurtz]

Thanks a bunch shashi_a_n_k_a_m

you are awesome!

And he is also violating forum rules by giving a complete solution.

 

[der.physika]

And he is also violating forum rules by giving a complete solution.

Okay, but I still don't know how to find that integral, which is what I'm going to go find out now. In the end I still learn something, even though I got the answer, I need to show my work completely, what's an answer without an explanation? a meaningless number...

 

[der.physika]

I figured out how to do that integral, the steps are as followed

(1) \int\frac{1}{1+\cos(u)}

(2) use \cos(u)\equiv1-\cos^2(\frac{u}{2})

and then it simplifies to

(3) \frac{1}{2}\sec^2(\frac{u}{2})

which comes out to

(4) \frac{1}{2}tan(\frac{x+y}{2})=y

 

[der.physika]

Isn't there a way I can write the equation in terms of only X values?