Second order DE from my midtearm

In summary, the function y=x^{2}, y=x and y\equiv1 are all solutions of the equation. However, you may not be able to derive x^{2} from the other two solutions. You may also be able to find a singular solution for y=1. The general solution is y(x)=c1*x+c2*x^2. However, you must choose Y_{p} to get the actual solution. Thanks for your help!
  • #1
Danny B
5
0

Homework Statement



Consider the ODE:

y''+p(x)y'+q(x)y=g(x)

It is given that the functions y=x[itex]^{2}[/itex], y=x and y[itex]\equiv[/itex]1
are solutions of the equation.
Find the general solution of the equation.

Homework Equations



The Attempt at a Solution



Well, given the three solutions, and taking into account that they are mutually linearly independent, I would say that any couple of these functions will give me a set that forms the general solution.

If i take y=x and y[itex]\equiv[/itex]1, I get the general solution:

y(x)=c1+c2*x

Now, I know that x[itex]^{2}[/itex] is also a solution, but it can't be derived from the general solution, so obviously I'm doing something wrong.

I also thought of the possibility that y=1 is a singular solution and the general solution is:

y(x)=c1*x+c2*x^2

does this makes sense?

This is a question from my midterm and i still can't figure it out. So... help?
 
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  • #2
Welcome to PF!

Hi Danny B! Welcome to PF! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)

Each of those three are particular solutions …

Hint: what relation is there between any two particular solutions (ie any two solutions)? :wink:
 
  • #3
Thank you.

I think I got it now.

My mistake was to use the principle of superposition with a non-homogeneous equation.

So the general solution should be:

(Considering that any difference of the particular solutions is a homogeneous solution)

y[itex]_{(x)}[/itex]=c[itex]_{1}[/itex](x-1)+c[itex]_{2}[/itex](x[itex]^{2}[/itex]-1)+Y[itex]_{p}[/itex]

But how do I choose Y[itex]_{p}[/itex]?

Can I just use any of the three particular solutions?
 
  • #4
Hi Danny B! :smile:

Let's write it out in full, so you can see how it works …

the general solution can be written in three ways

A(x2 - 1) + B(x - 1) + 1

A(x2 - x) + x + B(1 - x)

x2 + A(x - x2) + B(1 - x2)​

the homogeneous general solution can be written in three ways

A(x2 - 1) + B(x - 1)

A(x2 - x) + B(1 - x)

A(x - x2) + B(1 - x2)​

now you need to spend ten minutes convincing yourself that they're the same! :rolleyes:

he he :biggrin:
 
  • #5
Thanks a lot!

You've been a great help!
 

FAQ: Second order DE from my midtearm

1. What is a second order differential equation?

A second order differential equation is a mathematical equation that describes the relationship between a function and its second derivative. It is written in the form of y'' = f(x, y, y') where y' represents the first derivative and y'' represents the second derivative of the function y with respect to the independent variable x.

2. How are second order differential equations solved?

Second order differential equations can be solved using a variety of methods such as separation of variables, substitution, and the method of undetermined coefficients. The choice of method depends on the form and complexity of the equation.

3. What are the applications of second order differential equations?

Second order differential equations are used in many fields of science and engineering, including physics, chemistry, biology, and economics. They are particularly useful in modeling systems with acceleration, such as motion, oscillations, and electrical circuits.

4. What is the difference between a homogeneous and non-homogeneous second order differential equation?

A homogeneous second order differential equation has a zero on the right side of the equation, while a non-homogeneous equation has a non-zero function on the right side. Homogeneous equations have a general solution that can be found using the methods mentioned above, while non-homogeneous equations require an additional particular solution to be added to the general solution.

5. Can all second order differential equations be solved analytically?

No, not all second order differential equations have analytical solutions. Some equations are too complex to be solved analytically and require numerical methods or approximation techniques to find a solution. Additionally, some equations may not have a general solution and only have particular solutions for specific initial conditions.

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