I have already solved the following non-homogeneous, second order differential equation:


I am dissatisfied by the way I found the particular solution. I assumed that the particular solution would be of the form $y(x)=a+bx+cx^2$, put that into the LHS and then compared coefficients.

  • How may I derive the particular solution, i.e. find that the particular solution is $1+bx+x^2$ without assuming it is of the form $a+bx+cx^2$ and then showing $a=1$ and $c=1$?

Remark: Amzoti's comment "OP does not want to try a different approach, so removed" to his deleted post is untrue and misleading. He suggested a substitution. The given substitution didn't even find the complementary solution, nevermind explain the particular solution.


There are two common methods for finding the particular solution of a second order non-homogenous ODE; the method of undetermined coefficients and variation of parameters. It seems as if you have chosen the former as one of the steps requires starting with a linear combination of the term(s) that make up the forcing function (right hand side). In your case, $a + bx + cx^{2}$. This is the easier of the two methods for equations where certain necessary conditions are met. However, the method of variation of parameters is more general and does not assume the functional form to be a linear combination of the rhs. There is gobs of literature written about this method, and I'm not certain this is what you're interested in, but if so, here is one link to take a peek:


Hope this helps some.


Paul Safier


I'm going to present a justification for the choice of the particular solution and its relation with the input function $g(x)$ - RHS; and then I'm going to discuss the case you provided.

Note that the general method of undetermined coefficients is limited to nonhomogeneous linear DE with constant coefficients. Since the input function $g(x)$ results from applying the polynomial differential operator $L$ to the particular solution $y_p$; it's reasonable for the input function to equal a linear combination consisting of $y_p$ and its derivatives---by a linear combination I mean multiplying each term by a constant and adding the results. This is possible because the coefficients of the DE are constants. The method is exclusive for the cases where $g(x)$ is a polynomial, an exponential, a sine, a cosine or an arithmetic combination of them by means of addition, substraction or/and multiplication.

The DE you provided is with noncostant coefficients, it would be wrong to apply the method of undetermined coefficients in this case. However, as a counter-example let's try to use the method here. $$(x + 1)y'' + xy' - y = x^2 + 2x + 1.$$

According to the method we assume:

$$y_p = Ax^2 + Bx + C,$$ so we have $y_p' = 2Ax + B$ and $y_p'' = 2A$. We substitute them into the DE to obtain: $$(x + 1)\cdot 2A + x(2Ax + B) - Ax^2 - Bx - C = x^2 + 2x + 1,$$ $$Ax^2 + 2Ax + (2A - C)= x^2 + 2x + 1.$$ As you can see, when we compare the coefficients of $x$ on both sides to figure out the values of $A, B$ and $C$; we find that $B$ is dropped in the LHS and we can never find its value.

**Note the difference between my substitution and yours: I used $A$ instead of $c$ and $C$ instead of $a$, it's a matter of habit of course.

I couldn't fully understand your last sentence. If you are dissatisfied with the process of assuming the coefficients and comparing LHS with RHS to find their values, you can use the method of variation of parameters. However, variation of parameters requires that you know the complementary function $y_c$ first.

Because the coefficients of the given equation are noncostant, you need to use some substitution or the Laplace Transform in order to solve the associated homogeneous DE.

I hope my answer sheds some light on your problem. Kind regards, Khalid Adel.


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