Substitution in Differential Equations

[rgalvan2]

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))
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!

 

[rootX]

f =

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))

 

[Mark44]

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))

Everything is fine to here, but goes downhill after that.
The integral on the right isn't too bad.

$$\int \frac{dv}{1 - sin(v)}$$
$$= \int \frac{dv}{1 - sin(v)} * \frac{1 + sin(v)}{1 + sin(v)}$$
$$=\int \frac{(1 + sin(v))dv}{1 - sin^2(v)}$$

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!