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Step-by-Step Differential Equation Solutions in Wolfram|Alpha

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❶It is now time to see why having the complementary solution in hand first is useful. Everywhere we see a product of constants we will rename it and call it a single constant.
## The Method of Undetermined Coefficients

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Everywhere we see a product of constants we will rename it and call it a single constant. This is a general rule that we will use when faced with a product of a polynomial and a trig function. We write down the guess for the polynomial and then multiply that by a cosine. We then write down the guess for the polynomial again, using different coefficients, and multiply this by a sine. This final part has all three parts to it.

First, we will ignore the exponential and write down a guess for. Writing down the guesses for products is usually not that difficult. The difficulty arises when you need to actually find the constants. This fact can be used to both find particular solutions to differential equations that have sums in them and to write down guess for functions that have sums in them.

There is nothing to do with this problem. All that we need to do it go back to the appropriate examples above and get the particular solution from that example and add them all together.

This is in the table of the basic functions. However, we wanted to justify the guess that we put down there.

Using the fact on sums of function we would be tempted to write down a guess for the cosine and a guess for the sine. So, we would get a cosine from each guess and a sine from each guess. The problem with this as a guess is that we are only going to get two equations to solve after plugging into the differential equation and yet we have 4 unknowns. We will never be able to solve for each of the constants. The general rule of thumb for writing down guesses for functions that involve sums is to always combine like terms into single terms with single coefficients.

This will greatly simplify the work required to find the coefficients. For this one we will get two sets of sines and cosines. This will arise because we have two different arguments in them.

The main point of this problem is dealing with the constant. We just wanted to make sure that an example of that is somewhere in the notes. If you recall that a constant is nothing more than a zeroth degree polynomial the guess becomes clear.

All we did was move the 9. However, upon doing that we see that the function is really a sum of a quadratic polynomial and a sine. The guess for this is then. Notice however that if we were to multiply the exponential in the second term through we would end up with two terms that are essentially the same and would need to be combined.

This is a case where the guess for one term is completely contained in the guess for a different term. Therefore, we will take the one with the largest degree polynomial in front of it and write down the guess for that one and ignore the other term. So, the guess for the function is. This last part is designed to make sure you understand the general rule that we used in the last two parts. This time there really are three terms and we will need a guess for each term.

The guess here is. We can only combine guesses if they are identical up to the constant. So, when dealing with sums of functions make sure that you look for identical guesses that may or may not be contained in other guesses and combine them.

This will simplify your work later on. We have one last topic in this section that needs to be dealt with.

In the first few examples we were constantly harping on the usefulness of having the complementary solution in hand before making the guess for a particular solution. It is now time to see why having the complementary solution in hand first is useful. This problem seems almost too simple to be given this late in the section. There is not much to the guess here. From our previous work we know that the guess for the particular solution should be,. Something seems wrong here. So, what went wrong?

We finally need the complementary solution. Notice that the second term in the complementary solution listed above is exactly our guess for the form of the particular solution and now recall that both portions of the complementary solution are solutions to the homogeneous differential equation,.

In other words, we had better have gotten zero by plugging our guess into the differential equation, it is a solution to the homogeneous differential equation! So, what did we learn from this last example. This work is avoidable if we first find the complementary solution and comparing our guess to the complementary solution and seeing if any portion of your guess shows up in the complementary solution.

In order to give the complete solution of a nonhomogeneous linear differential equation, Theorem B says that a particular solution must be added to the general solution of the corresponding homogeneous equation.

The special functions that can be handled by this method are those that have a finite family of derivatives, that is, functions with the property that all their derivatives can be written in terms of just a finite number of other functions. Here's an example of a function that does not have a finite family of derivatives: Its first four derivatives are. Note that any numerical coefficients such as the 5 in this case are ignored when determining a function's family.

The central idea of the method of undetermined coefficients is this: Find a particular solution of the differential equation.

Substituting this into the given differential equation gives. Now, combinbing like terms yields. The first equation immediately gives. Therefore, a particular solution of the given differential equation is. Find a particular solution and the complete solution of the differential equation. Now, combining like terms and simplifying yields. Posted by Sam Blake January 30, at Posted by Brian Gilbert January 31, at 6: Posted by Oriol February 1, at 3: Posted by Bhuvanesh February 3, at 8: Posted by rymo February 7, at 3: Thanks for a great piece of tech.!

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Solve integrals with Wolfram|Alpha. Step-by-step Solutions» Walk through homework problems step-by-step from beginning to end. Hints help you try the next step on your own. Wolfram Problem Generator» Unlimited random practice problems and answers with built-in Step-by-step solutions. Practice online or make a printable study sheet.

Wolfram|Alpha can help out in many different cases when it comes to differential equations. Get step-by-step directions on solving exact equations or get help on solving higher-order equations. Even differential equations that are solved with initial conditions are easy to compute.

Pros and Cons of the Method of Undetermined Coefficients:The method is very easy to perform. However, the limitation of the method of undetermined coefficients is that the non-homogeneous term can only contain simple functions such as,,, and so the trial function can be effectively guessed. I am having trouble figuring out how to find a particular solution to a differential equation using the method of undetermined coefficients. Everything I've found from other sites hasn't worked.

The central idea of the method of undetermined coefficients is this: Form the most general linear combination of the functions in the family of the nonhomogeneous term d(x), substitute this expression into the given nonhomogeneous differential equation, and solve for . The class of \(g(t)\)’s for which the method works, does include some of the more common functions, however, there are many functions out there for which undetermined coefficients simply won’t work. Second, it is generally only useful for constant coefficient differential equations. The method is .