# How to Solve a Second Order Nonhomogeneous Differential Equation?

• Math10
In summary: If so, what is the solution?Yes, I checked and the supposed answer solves the DE. The solution is yp=e^(x)(x^2*cosx+2sinx).
Math10

## Homework Statement

Find a particular solution of y"-5y'+6y=-e^(x)[(4+6x-x^2)cosx-(2-4x+3x^2)sinx].

None.

## The Attempt at a Solution

yp=e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
y'p=e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
y"p=e^(x)[(Ax^2+Bx+C)(-cosx+sinx)+(2Ax+B)(-sinx-cosx)+(2Ax+B)(-sinx-cosx)+2A(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
subbing:
y"p-5y'p+6yp=4Ax^2e^xcosx+4Bxe^xcosx+4Ce^xcosx+2Ax^2e^xsinx+2Bxe^xsinx+2Ce^xsinx+2Axe^xsinx+Be^xsinx-10Axe^xcosx-5Be^xcosx+2Ae^xcosx-2Ae^xsinx=-e^x[(4+6x-x^2)cosx-(2-4x+3x^2)sinx]
So 4Ax^2e^xcosx=x^2e^xcosx
and therefore A=1/4.
But the answer to this problem is yp=e^(x)(x^2*cosx+2sinx). I don't think the coefficients I got are right. So what's the right initial guess for this problem?

How did you come up with your guess?

Just by making my best guess.

What made you think it was the best as opposed to some other form? I'm asking because writing down the form of particular solution is a cookbook process, yet you managed to come up with something different.

Since there's -e^x in the very front, so I got e^x as well in forming my guess yp. And then there's (4+6x-x^2) in front of cosx and same thing in front of sinx. So I got yp=e^(x)[(Ax^2+Bx+C)(cosx-sinx)].

The sine doesn't have the same factor multiplying it as cosine in the forcing function, or are you saying the cosine and the sine should both have a quadratic multiplying it? If the latter, why would you expect it to be the exact same quadratic?

So this is a forcing function?

The forcing function is the stuff that appears on the righthand side of the differential equation.

So what's the way to find the guess?

Math10 said:

## Homework Statement

Find a particular solution of y"-5y'+6y=-e^(x)[(4+6x-x^2)cosx-(2-4x+3x^2)sinx].

None.

## The Attempt at a Solution

yp=e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
y'p=e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
y"p=e^(x)[(Ax^2+Bx+C)(-cosx+sinx)+(2Ax+B)(-sinx-cosx)+(2Ax+B)(-sinx-cosx)+2A(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
subbing:
y"p-5y'p+6yp=4Ax^2e^xcosx+4Bxe^xcosx+4Ce^xcosx+2Ax^2e^xsinx+2Bxe^xsinx+2Ce^xsinx+2Axe^xsinx+Be^xsinx-10Axe^xcosx-5Be^xcosx+2Ae^xcosx-2Ae^xsinx=-e^x[(4+6x-x^2)cosx-(2-4x+3x^2)sinx]
So 4Ax^2e^xcosx=x^2e^xcosx
and therefore A=1/4.
But the answer to this problem is yp=e^(x)(x^2*cosx+2sinx). I don't think the coefficients I got are right. So what's the right initial guess for this problem?

Have you checked whether the supposed "answer" actually solves the DE?

Last edited:

## What is a particular solution?

A particular solution is a specific solution to a differential equation that satisfies all of the given initial conditions. It is used to find the general solution to a differential equation.

## How do you find a particular solution?

To find a particular solution, you first need to identify the type of differential equation and any given initial conditions. Then, you can use various methods such as separation of variables, integration, or substitution to solve for the particular solution.

## Why is finding a particular solution important?

Finding a particular solution allows you to completely solve a differential equation and obtain a general solution. This is important because it helps to understand the behavior of a system or process that is governed by the differential equation.

## What if I cannot find a particular solution?

If you are unable to find a particular solution, it is possible that the differential equation may not have a solution or that you made an error in your calculations. In this case, it is important to check your work and consult with a mentor or colleague for further assistance.

## Can a particular solution be unique?

Yes, a particular solution can be unique. This means that there is only one solution that satisfies all of the given initial conditions. However, in some cases, a particular solution may not be unique and there may be multiple solutions that satisfy the given conditions.

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