Understanding DE's: Solving for y=x and Exam Prep Tips

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In summary, the "integrating factor" site Hootenanny gave you to solve a first order equation for u to find v which will then give you y= xv.
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franky2727
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I have a question of show that y=x is one solution of the equation (1+x^2)y''-2xy'+2y=0) this is a question that i need to know for a coming exam. i understand and can do everything until the question gets to v''+(2/x(1+x^2))v'=0

then the question skips to IF =e^integral of (2/x(1+x^2))dx. what does this mean and where does it come from
 
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  • #2
franky2727 said:
I have a question of show that y=x is one solution of the equation (1+x^2)y''-2xy'+2y=0) this is a question that i need to know for a coming exam. i understand and can do everything until the question gets to v''+(2/x(1+x^2))v'=0

then the question skips to IF =e^integral of (2/x(1+x^2))dx. what does this mean and where does it come from
IF is known as the http://en.wikipedia.org/wiki/Integrating_factor" [Broken] as it one technique used to solve ODE's.
 
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  • #3
ok thanks and why is the integrating factor that part of the equation?
 
  • #4
franky2727 said:
ok thanks and why is the integrating factor that part of the equation?
The linked article explains why under the heading "Technique".

Do you follow, or does it need more explanation?
 
  • #5
franky2727 said:
I have a question of show that y=x is one solution of the equation (1+x^2)y''-2xy'+2y=0) this is a question that i need to know for a coming exam. i understand and can do everything until the question gets to v''+(2/x(1+x^2))v'=0

then the question skips to IF =e^integral of (2/x(1+x^2))dx. what does this mean and where does it come from
If the problem is, as you say, just to show that y= x is one solution then you don't need to do all that. If y= x, then y'= 1 and y"= 0 so the equation becomes
(1+ x2)(0)- 2x(1)+ 2(x)= -2x+ 2x= 0. Q.E.D.

However, I suspect that the problem asks you to (1) show that y= x is a solution to the equation and then (2) find another, independent, solution.

Since y= x is a solution, we can look for another solution of the form y= xv(x) where v(x) is an unknown function of x. Now y'= xv'+ v and y"= xv"+ 2v'. Putting those into the equation, (1+ x^2)(xv"+ 2v')- 2x(xv'+ v)+ 2xv= (x+ x^3)y"+ (2+ 2x^2- 2x^2)v' - 2xv+ 2xv= (x+ x^3)v"+ 2v'= 0. While that is a second order equation, v itself does not appear in the equation (because y= x satisfies the equation) and, letting u= v', we have a first order equation, (x+ x^3)u'+ 2u= 0. Now you can use the "integrating factor" site Hootenanny gave to solve that first order equation for u, then integrate to find v and, finally, find y= xv.
 

1. What is the difference between an ordinary differential equation (ODE) and a partial differential equation (PDE)?

An ODE involves a single independent variable and its derivatives, while a PDE involves multiple independent variables and their derivatives.

2. How do I solve a differential equation where y=x?

To solve for y=x, you can simply substitute x for y in the differential equation and solve for x. This will result in the solution y=x.

3. What are some tips for preparing for exams on differential equations?

Some tips for exam preparation include practicing solving different types of DEs, reviewing key concepts and formulas, and working through past exam questions or practice problems.

4. How do I know which method to use when solving a differential equation?

The method used to solve a differential equation depends on its type and order. Some common methods include separation of variables, substitution, integrating factors, and power series. It is important to identify the type of DE and choose the appropriate method.

5. Can differential equations be used in real-life applications?

Yes, differential equations are widely used in many fields of science and engineering to model real-life situations. They can be used to describe the change of physical quantities over time and are essential in understanding and predicting various phenomena in nature.

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