I am getting stuck on finding a particular solution to the equation:


I know the general solution must be of the form $y_g=y_p+y_h$

I fround the corresponding homogeneous solution to be $y_h=c_1e^{3x}+c_2xe^{3x}$ and I am stuck trying to find a particular solution $y_p$. The method that should be used is the method of undetermined coefficients, and I have attempted solutions of the form $Axe^x, Ae^x+(Bx+c)e^{3x}$, and actually a few others to no avail. If someone could tell me the solution and $how$ they determined what form was necessary for the particular solution I would be much obliged.


$(D-3)^2[y]=(x+1)e^{3x}$ acted on by $(D-3)^2$ yields $(D-3)^4[y]=0$ which suggests $y_p = Ax^3e^{3x}+Bx^2e^{3x}$. Your homogeneous solution is fine.

I'm using the method of annihilators which works for differential equations with constant coefficients where the inhomogeneous term is something which can arise in the solution set of a constant coefficient ODE. Generically, a constant coefficient ODE $y^{n}+a_{n-1}y^{n-1}+ \cdots + a_1y'+a_0y = g$ can be written in operator notation as $(D^n+a_{n-1}D^{n-1} + \cdots + a_1D+a_0)[y] = g$ where $D = d/dx$. If you can find another polynomial in $D$ which kills $g$ then this $A$ is said to annihilate $g$; $A[g]=0$. Thus, $$ (D^n+a_{n-1}D^{n-1} + \cdots + a_1D+a_0)[y] = g \ \ \Rightarrow \ \ A(D^n+a_{n-1}D^{n-1} + \cdots + a_1D+a_0)[y] = A[g]=0 $$ Let's call $D^n+a_{n-1}D^{n-1} + \cdots + a_1D+a_0 = L$ so we have $AL[y]=0$ and this is in fact a constant coefficient ODE so we can find the general solution. On the other hand, you know $y_p$ solves $L[y_p]=g$ hence $y_p$ must be covered by the general solution to $AL[y]=0$ (note $AL[y_p]=A[g]=0$ so it's in there). Thus, the form for $y_p$ can be ciphered from $AL[y]=0$. In practice, you find two parts, the homogeneous solution and $y_p$ the both appear in the $AL[y]=0$ solution. This is known as the method of annhilators

  • $\begingroup$ What theorem/result is being used because I was attempting something totally different and it would be helpful to know. $\endgroup$ – wesssg Oct 4 '17 at 3:22
  • 1
    $\begingroup$ youtube.com/… might be helpful for you. $\endgroup$ – James S. Cook Oct 4 '17 at 17:39

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