- #1

- 30

- 0

Here's what I did:

v=x-y

y=x-v

y'=1-dv/dx

1-dv/dx=sin(v)

1-sin(v)=dv/dx

dx=dv/(1-sin(v))

x=2/(cot(v/2)-1)

The solution in the back of the book gives:

x=tan(x-y) + sec(x-y)

What am I doing wrong?

Any help is greatly appreciated.

Thanks!

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- Thread starter rgalvan2
- Start date

- #1

- 30

- 0

Here's what I did:

v=x-y

y=x-v

y'=1-dv/dx

1-dv/dx=sin(v)

1-sin(v)=dv/dx

dx=dv/(1-sin(v))

x=2/(cot(v/2)-1)

The solution in the back of the book gives:

x=tan(x-y) + sec(x-y)

What am I doing wrong?

Any help is greatly appreciated.

Thanks!

- #2

- 412

- 2

1/(1-sin(x))

>> int(f,x)

ans =

-2/(tan(1/2*x)-1)

I have something different from matlab when I integrate 1/(1-sin(x))

- #3

Mark44

Mentor

- 35,125

- 6,872

Everything is fine to here, but goes downhill after that.Make an appropriate substitution to find a solution of the equation dy/dx=sin(x-y). Does this general solution contain the linear solution y(x)=x-pi/2 that is readily verified by substitution in the differential equation?

Here's what I did:

v=x-y

y=x-v

y'=1-dv/dx

1-dv/dx=sin(v)

1-sin(v)=dv/dx

dx=dv/(1-sin(v))

The integral on the right isn't too bad.

[tex]\int \frac{dv}{1 - sin(v)}[/tex]

[tex]= \int \frac{dv}{1 - sin(v)} * \frac{1 + sin(v)}{1 + sin(v)}[/tex]

[tex]=\int \frac{(1 + sin(v))dv}{1 - sin^2(v)}[/tex]

The denominator simplifies to cos^2(v) and you can split the integral into two integrals, one of which is straightforward. The other one requires only an ordinary substitution.

I ended with the same answer as in the book.

x=2/(cot(v/2)-1)

The solution in the back of the book gives:

x=tan(x-y) + sec(x-y)

What am I doing wrong?

Any help is greatly appreciated.

Thanks!

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