Don't know how to solve equation

[Incubator]

Hi, i'm having problem with an equation...
The exams are soon, and i don't know how to start solving this problem
I would greatly appreciate any help


Problem:
y = x y' - cos y'

 

[Deleted member 4993]

Hi, i'm having problem with an equation...
The exams are soon, and i don't know how to start solving this problem
I would greatly appreciate any help


Problem:
y = x y' - cos y'

This is famous Clairaut's equation.

As a first step, differentiate both sides again wrt 'x'.

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[Incubator]

Actually this is where i'm stuck...

i started by y' = t
then y = x*t - cos(t)

and here i should differentiate both sides with respect to x
which i believe would be something like this ie. in a simple equation:
x^2 + 4 + y = cos(x) + x*y^2 + 4x^2
=> 2x = -sin(x) + y^2 + 8x

but in my equation that would mean 0 = t which i know is wrong :\

 

[HallsofIvy]

Actually this is where i'm stuck...

i started by y' = t
then y = x*t - cos(t)

and here i should differentiate both sides with respect to x
which i believe would be something like this ie. in a simple equation:
x^2 + 4 + y = cos(x) + x*y^2 + 4x^2
=> 2x = -sin(x) + y^2 + 8x

but in my equation that would mean 0 = t which i know is wrong :

That makes no sense at all. What do the equations at the bottom have to do the original equation? Also it makes no sense to me to introduce a new variable for the derivative of y. (And the last equation is NOT the first equation differentiated with respect to x.)

Your original equation is y = x y' - cos y' and differentiating again with respect to x gives y'= y'+ xy''- sin(y')y'' or sin(y')y''= xy''.

 

[Incubator]

Is this correct?

y = x y' - cos(y')
y' = t

y = x t - cos(t)
dy = t dx + (x + sin(t) ) dt

(x + sin(t) ) dt = 0

If
dt = 0, t = c
=> y = cx - cos(c)

If
x + sin(t) = 0
then
x = -sin(t)
y = xt - cos(t)
y = t sin(t) - cos(t)

 

[HallsofIvy]

Is this correct?

y = x y' - cos(y')
y' = t

y = x t - cos(t)
dy = t dx + (x + sin(t) ) dt

(x + sin(t) ) dt = 0

If
dt = 0, t = c
=> y = cx - cos(c)

If
x + sin(t) = 0
then
x = -sin(t)
y = xt - cos(t)
y = t sin(t) - cos(t)

Did you check those solutions in youroriginal equation? (The second, y= t sin(t)- cos(t), clearly is NOT a solution because y is a function of x, not t. t is a variable you introduced into the problem.

 

[Lost souls]

Is this correct?

y = x y' - cos(y')
y' = t

y = x t - cos(t)
dy = t dx + (x + sin(t) ) dt

(x + sin(t) ) dt = 0

If
dt = 0, t = c
=> y = cx - cos(c)

If
x + sin(t) = 0
then
x = -sin(t)
y = xt - cos(t)
y = t sin(t) - cos(t)

y= xy' - cos(y'). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(1)

differentiating wrt "x",

=> y' = y' + xy" + sin(y').y"


cancelling y' from both sides,

=> xy" + sin(y').y" = 0

=> {x+sin(y')}y" = 0

=> y"= 0

integrating once,

=> y' = C (where C is a constant)

putting value of y' in eqn(1)

y = xC - cos(C). . . . . . . . . . .is the required solution.

 

[HallsofIvy]

y= xy' - cos(y'). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(1)

differentiating wrt "x",

=> y' = y' + xy" + sin(y').y"


cancelling y' from both sides,

=> xy" + sin(y').y" = 0

=> {x+sin(y')}y" = 0

=> y"= 0

This does not follow. What you can conclude is that either y''= 0 or x+ sin(y')= 0 which leads to y'= arcsin(-x)= -arcsin(x).

integrating once,

=> y' = C (where C is a constant)

putting value of y' in eqn(1)

y = xC - cos(C). . . . . . . . . . .is the required solution.