How to solve these Differential Equations?

In summary: So My - Nx = \frac{ye^{\frac{y}{x}}}{x} + \frac{2x^3}{(1 + x^2)^2} + \frac{ye^{\frac{y}{x}}}{x^2} - \frac{2x^3}{(1 + x^2)^2}= \frac{ye^{\frac{y}{x}}(x+1
  • #1
Naeem
194
0
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.
 
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  • #2
The second one is really simple...

[tex]\frac{dx}{dy}=\frac{ye^{y}-2x}{y} [/tex]

Nonhomogenous linear ODE.The homogenous one is separable.

Daniel.
 
  • #3
Alright, let's write these out:

1. [tex]\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[/tex]

2. [tex]y \ dx + (2x - ye^y) \ dy = 0[/tex]

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

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

An integrating factor that will work for #2 is just [itex]y[/itex] (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...).
 
  • #4
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
 
  • #5
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
 
  • #6
[tex]\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[/tex]

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

[tex] 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}[/tex]

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

Related to How to solve these Differential Equations?

1. How do I know which method to use when solving a Differential Equation?

There are various methods for solving Differential Equations, such as separation of variables, substitution, and integrating factors. The method you should use depends on the type of Differential Equation and its variables. It is important to understand the characteristics of each method and how they apply to the equation you are solving.

2. Can I solve a Differential Equation by hand or do I need a computer?

Some simple Differential Equations can be solved by hand using algebraic manipulation and integration techniques. However, as the equations become more complex, it may be necessary to use a computer or specialized software to find the solution. This is especially true for systems of Differential Equations.

3. How do I check if my solution to a Differential Equation is correct?

One way to check the solution to a Differential Equation is to substitute the solution into the original equation and see if it satisfies the equation. Another method is to plot the solution and compare it to the graph of the original equation. Additionally, you can use numerical methods and compare the results to your solution.

4. Can I use the same method for all types of Differential Equations?

No, different types of Differential Equations require different methods for solving them. For example, you cannot use the same method for solving a linear Differential Equation as you would for a nonlinear one. It is important to understand the properties of each type of Differential Equation and choose the appropriate method for solving it.

5. Are there any tricks or shortcuts for solving Differential Equations?

While there are no shortcuts for solving all types of Differential Equations, there are some techniques that can make the process easier. For instance, in some cases, you can transform a higher-order Differential Equation into a system of first-order equations, which can be easier to solve. It is also helpful to become familiar with common types of equations and their solutions, as this can make it easier to recognize the appropriate method to use.

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