Differential equation from general solution

In summary, the conversation discusses finding the differential equation for a given general solution, using the example of y=sin(ax+b). The method involves finding the first and second derivatives of the function, and then manipulating them algebraically to eliminate the parameters and result in a second order differential equation. This can be done by guessing a solution and solving for the characteristic equation, or by using the general solution for a diff eq with complex roots. The conversation concludes with the acknowledgement that the method worked successfully in this case.
  • #1
Damidami
94
0
What if I have the general solution and I want to find the differential equation where it came from?

Say for example my general solution is [tex]y=sin(ax + b)[/tex] with [tex]a[/tex] and [tex]b[/tex] constants.
How could I find the differential equation?
 
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  • #2
I don't know any particular pattern, so i am just guessing. Try to find say the first and second derivative of this function.And after that see if you can find any linear combination equal to zero,or to sth else, of second, first and the originall function
 
  • #3
Like [tex] y'=acos(ax+b)[/tex]

[tex] y=sin(ax+b)[/tex]

[tex] y''=-a^2sin(ax+b)[/tex]

Now one can easily notice that

[tex] y''+a^2y=0[/tex] so i guess this would be one diff eq whose solution would be

[tex] y=sin(ax+b)[/tex]
 
  • #4
Now let's try to solve this diff equation


[tex]y''+a^2y=0[/tex] and pretend that we actually don't know the solution to it. so

let's guess a solution [tex]y=e^{rx}[/tex]
[tex]y''=r^2e^{rx}[/tex] so the char. eq of that diff. eq is

[tex]r^2+a^2=0=>r^2=-a^2=>r=\pm a i[/tex]

We know that the general solution of a diff eq. with complex roots is:

[tex]y=e^{\alpha x}(c_1cos(\beta x)+c_2sin(\beta x))[/tex]

Now we have [tex] \alpha =0, \beta=a [/tex] so the general solution to our diff. eq would be:

[tex]y=e^{x0}(acos(ax)+bsin(ax))=c_1cos(ax)+c_2sin(ax)[/tex] Which is the same as our original answer;

[tex] y=sin(ax+b)=sin(ax)cosb+cos(ax)sinb=c_1cos(ax)+c_2sin(ax)[/tex]

notice that

[tex] cos(b)=c_1, sinb=c_2[/tex] these are constants.

Wow, it worked out! I'm glad...lol...
 
  • #5
Basically your objective is to eleminate the parameters, here a and b, by differentiating and algebraically manipulating the equations. Typically, you will need as many derivatives as you have parameters to do that. Since you have two parameters here you would expect to result in a second order differential equation.

Sutupid Math did that quite nicely for you.
 

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model many physical, chemical, and biological phenomena in the natural world.

2. What is a general solution to a differential equation?

A general solution to a differential equation is an equation that contains all possible solutions to the differential equation. These solutions are represented by a constant, known as the arbitrary constant, which can take on different values for different specific solutions.

3. How is a general solution different from a particular solution?

A particular solution to a differential equation is a specific solution that satisfies both the differential equation and any initial conditions. It is obtained by substituting specific values for the arbitrary constant in the general solution.

4. What is the process for solving a differential equation using the general solution?

The process for solving a differential equation using the general solution involves finding the general solution, which typically involves integrating the equation and including an arbitrary constant. Then, specific values are substituted for the arbitrary constant to find particular solutions that satisfy any given initial conditions.

5. Can all differential equations be solved using the general solution?

No, not all differential equations can be solved using the general solution. Some equations are too complex and do not have a known general solution. In these cases, numerical methods or approximations are used to find solutions.

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