Diff Equation problem- give me a nudge, please

[Latecomer]

Homework Statement

(dy/dx) - cos²(x-y) = 0

Homework Equations

The Attempt at a Solution

I'm unsure where to start this. This isn't a liner equation of form (dy/dt) + p(t)y = g(t) ; and it's not separable nor an exact equation.

Is it illegal to make this (dy/dx) - cos²(x) + cos²y = 0 ?


Thanks for your time.

 

[ehild]

Is it illegal to make this (dy/dx) - cos²(x) + cos²y = 0 ?

It is totally illegal!

Hint: replace y-x with a new variable.

ehild

 

[Latecomer]

Yeah, I knew that was illegal. I was just grasping at straws because I had no idea where to begin. And it's early...

Well, I have not worked with problems like this at all and I'm having trouble finding anything similar in my text, but this is what I have figured out so far:

(dy/dx) -cos²(x-y) = 0

(dy/dx) = cos²(x-y) make (x-y) = v

(dy/dx) = cos²(v)

v = x - y
y = x - v
y' = 1 - v'

from original : y' = cos²(x - y)
so: 1 - v' = cos²(v)
and : v' = 1 - cos²(v)

Am I heading in the right direction? Another nudge? Thanks again.

 

[ehild]

It is all right so far. Is not it a separable de?

ehild

 

[Char. Limit]

Remember that 1-cos²(v) = sin²(v).

 

[Latecomer]

Yes, I should have seen that.

(dv/dx) = 1-cos²(v)

dv/1-cos²(v) = dx

which is : csc^²(v) = dx

integrate:

-cot(v) = x + c

-cot (x - y) = x + c

Thank you for your help. I was just staring dumbly at it.

This implicit form should be a suitable answer for this question, yes? I was just told to solve the equation.

 

[Char. Limit]

It should be suitable, but it's a simple process to get the answer in explicit form, if you want it.

 

[Latecomer]

Hmm...

-cot (x-y) = x + c

cot (x-y) = -x +c

(x - y) = arccot (-x + c)

-y= arccot (-x + c) - x

y = x - arccot (-x + c) ?

 

[Char. Limit]

Hint: The inverse of the cotangent function is arccotangent, not arctangent.

 

[Latecomer]

hehe, yeah I saw that as soon as I submitted it and then edited it. You're fast

 

[Char. Limit]

Looks good to me. You can plug it in and test it if you want.