How to solve these Differential Equations?

[Naeem]

Hi,

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.

 

[dextercioby]

The second one is really simple...

\frac{dx}{dy}=\frac{ye^{y}-2x}{y}

Nonhomogenous linear ODE.The homogenous one is separable.

Daniel.

 

[Data]

Alright, let's write these out:

1. \left(e^{\frac{y}{x}} - \frac{y}{x}e^{\frac{y}{x}} + \frac{1}{1+x^2}\right) \ dx + e^{\frac{y}{x}} \ dy = 0

2. y \ dx + (2x - ye^y) \ dy = 0

3. (2x + \tan y) \ dx + (x - x^2\tan y) \ dy = 0

#1 is exact (check for yourself!). It isn't homogeneous.

An integrating factor that will work for #2 is just y (you can determine it by inspection...).

Integrating factor isn't that bad for #3 (although you will definitely know which function to integrate after you find it...).

 

[Naeem]

So, far had now luck with part 1,

If it is exact,

I tried to find partial derivatives with respect to y and x for both terms, and then applied

My - Nx / N

and the Integrating factor gets really messy...

I got 2 ,,, piece of cake.

3. if M = 2x + tany
then

My = sec^2y

N = x - x^2 tany

Nx = 1 -2xtany

Then,

Integrating factor = Nx - My / M

1-2xtany - sec^2y / 2x tany

Trig, identity,

sec^2y - tan^2y = 1
sec^2y = 1 + tan^2y

Plugging in this into: 1-2xtany - sec^2y / 2x tany , for sec^2y

we get,

1-2xtany -1 -tan^2y / 2x tany

After simplifying the above, I got:

Integrating factor as - tany

When I try to multiply both sides by -tany, the expression doesn't get any simpler, but infact becomes harder to evaluate,

Any ideas, on how this could be made simpler.

Thanks

 

[Naeem]

1. How the heck to find the partial for no. 1 with respect to x and y...

It is a pain. I would appreciate if somebody could help me with number 1 and 3


Thanks

 

[whozum]

\left(e^{\frac{y}{x}} - \frac{y}{x}e^{\frac{y}{x}} + \frac{1}{1+x^2}\right) \ dx + e^{\frac{y}{x}} \ dy = 0

I can't help you with solving the equation but the partials:

M_y = \frac{e^{\frac{y}{x}}}{x} - \frac{e^{\frac{y}{x}}}{x}+\frac{ye^{\frac{y}{x}}}{x} = \frac{ye^{\frac{y}{x}}}{x}

N_x = \frac{-ye^{y/x}}{x^2}