- #1

Naeem

- 194

- 0

Q 1. ( e ^ y/x - y/x e ^ y/x + 1 / 1 + x^2 ) dx + e ^ y/x dy = 0

Ans. I know its Homogeneous sub, y = ux,

then, dy / dx = u dx + x du

I did this, and got to the point,

e^u dx + 1 / ( 1 + x^2) dx + x. e^u du = 0

How can we separate this now?

Q 2. y dx + ( 2x - y e^y ) dy = 0

I think we can use exactness, here

M = y

My = 1

N = 2x - y e^y

Nx = 2

Not exact,

so,

Integrating Factor would be : e ^ Integral My - Nx / N

Is this right. Integrating factor is getting to complicated to multiply the eqn, with. Any better way of doing this.

Q 3. ( 2x + tany) dx + ( x - x^2 tany ) dy = 0

I think here too, exactness, may be used, but any better way, if possible.