Finding particular solution to differential equations

In summary: Then suppose for y''-9y'+14y my yp=Ax2e4xTheny'=2Axe4x+4Ax2e4xy''=2Ae4x+8Axe4x+8Axe4x+16Ax2e4xWhen you plug this all back into the original equation, equate the coefficients to solve for A, it doesn't work. Need more than just one constant.
  • #1
lillybeans
68
1
There are two questions that I am trying to solve on web assignment. The goal is to find a general form of a particular solution to each ODE. The question asks me to represent all constants in the solution using "P,Q,R,S,T..etc.", in that order.

1. y'''-9y''+14y'=x2
2. y''-9y'+14y=x2e4x

For the first one I wrote:
yp=Px3+Qx2+Rx

Second one:
yp=(Px2+Qx+R)e4x

Neither of the answers are correct, according to the computer. Where did I go wrong?

Thanks.
 
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  • #2
The answers are not correct because you're not solving the equations correctly... First do you understand that your equations are non-homogenous?
 
  • #3
Zondrina said:
The answers are not correct because you're not solving the equations correctly... First do you understand that your equations are non-homogenous?

Yes, of course, that's why I am asked to find a particular solution. Where did I go wrong? please let me know!
 
  • #4
Okay, so take for example your second question. Notice that your derivatives and original function have to add up to x2e4x?

So your general solution must have the form : y = A(x2e4x) where A is some constant number.

Now take the required derivatives and see if they satisfy your equation. ( You'll get a bunch of constants which you have to add together and solve for A for the particular solution you want ).
 
  • #5
Zondrina said:
Okay, so take for example your second question. Notice that your derivatives and original function have to add up to x2e4x?

So your general solution must have the form : y = A(x2e4x) where A is some constant number.

Now take the required derivatives and see if they satisfy your equation. ( You'll get a bunch of constants which you have to add together and solve for A for the particular solution you want ).

I don't quite understand why your guess for the particular solution has the form Ax2e4x as opposed to (Ax2+Bx+C)e4x. Shouldn't you use the general form of the second degree polynomial?
 
  • #6
A non-homogenous equation with constant coefficients has the general solution y = Ag(t) where A is a constant.
 
  • #7
Zondrina said:
A non-homogenous equation with constant coefficients has the general solution y = Ag(t) where A is a constant.

Then suppose for y''-9y'+14y my yp=Ax2e4x

Then

y'=2Axe4x+4Ax2e4x
y''=2Ae4x+8Axe4x+8Axe4x+16Ax2e4x

When you plug this all back into the original equation, equate the coefficients to solve for A, it doesn't work. Need more than just one constant.
 

1. How do you find the particular solution to a differential equation?

There are several methods for finding the particular solution to a differential equation, depending on the type of equation. One common method is the method of undetermined coefficients, which involves guessing a solution based on the form of the differential equation and then solving for the coefficients. Another method is variation of parameters, which involves finding a general solution to the homogeneous equation and then using a change of variables to find a particular solution. Finally, for linear equations with constant coefficients, the method of annihilators can be used.

2. Can you provide an example of finding a particular solution to a differential equation?

Sure, let's consider the differential equation y'' + 4y' + 3y = 12. Using the method of undetermined coefficients, we can guess a particular solution of the form y_p = Ax + B, where A and B are constants. Plugging this into the equation, we get 2A + 4Ax + 4B + 3Ax + 3B = 12. Equating coefficients, we can solve for A and B to get the particular solution y_p = 2x + 1.

3. What is the difference between a general solution and a particular solution to a differential equation?

A general solution to a differential equation is a solution that satisfies the equation for all possible values of the independent variable. This solution usually includes one or more arbitrary constants. A particular solution, on the other hand, is a specific solution that satisfies the differential equation for specific values of the independent variable. It does not contain any arbitrary constants.

4. Can you use a computer program to find the particular solution to a differential equation?

Yes, there are many computer programs and software packages that can solve differential equations and find particular solutions. Some commonly used programs include Mathematica, MATLAB, and Maple. These programs use numerical methods or symbolic computation to find solutions to differential equations.

5. Are there real-life applications for finding particular solutions to differential equations?

Yes, differential equations are used to model a wide range of physical phenomena in fields such as physics, engineering, and biology. Finding particular solutions to these equations can help us understand the behavior of systems and make predictions about future events. For example, particular solutions to differential equations can be used to model the spread of diseases, the motion of objects under the influence of forces, and the growth of populations over time.

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