MHB Using the annihilator method to solve an ODE

[MarkFL]

Recently I was sent an ODE with the instructions to solve using the annihilator method which I have not used in over 15 years. This is my working, and I was hoping for feedback to see if I have correctly and efficiently applied the method.

Here is my working:

We are given the ODE:

(1) $\displaystyle y''+3y'-10y=xe^x+x+1$

Since $\displaystyle (D-1)^2$ annihilates $\displaystyle xe^x$ and $\displaystyle D^2$ annihilates $\displaystyle x+1$ then:

$\displaystyle A\equiv (D(D-1))^2$

annihilates $\displaystyle xe^x+x+1$.

Therefore, applying $\displaystyle A$ to both sides of (1) yields:

$\displaystyle A[y''+3y'-10y]=A[xe^x+x+1]$

$\displaystyle (D(D-1))^2(D^2+3D-10)[y]=0$

(2) $\displaystyle (D(D-1))^2(D+5)(D-2)[y]=0$

The auxiliary equation associated with (2) is:

$\displaystyle r^2(r-1)^2(r+5)(r-2)=0$

which has the roots:

$\displaystyle r=-5,\,0,\,1,\,2$ where the roots $\displaystyle r=0,\,1$ are repeated roots, i.e., of multiplicity 2.

Hence, a general solution to (2) is:

(3) $\displaystyle y(x)=c_1+c_2x+c_3e^x+c_4xe^x+c_5e^{2x}+c_6e^{-5x}$

Now, recall that a general solution to (1) is of the form $\displaystyle y(x)=y_h(x)+y_p(x)$. Since every solution to (1) is also a solution to (2), then $\displaystyle y(x)$ must have the form displayed on the right-hand side of (3). But, we recognize that:

$\displaystyle y_h(x)=c_5e^{2x}+c_6e^{-5x}$

and so there must exist a particular solution of the form:

$\displaystyle y_p(x)=c_1+c_2x+c_3e^x+c_4xe^x$

In order to substitute this into (1), we must first compute:

$\displaystyle y_p'(x)=c_2+c_3e^x+c_4e^x(x+1)$

$\displaystyle y_p''(x)=c_3e^x+c_4e^x(x+2)$

and so we find:

$\displaystyle (c_3e^x+c_4e^x(x+2))+3(c_2+c_3e^x+c_4e^x(x+1))-10(c_1+c_2x+c_3e^x+c_4xe^x)=xe^x+x+1$

Collecting like terms, we may write:

$\displaystyle (-6c_4)xe^x+(5c_4-6c_3)e^x+(-10c_2)x+(3c_2-10c_1)=(1)xe^x+(0)e^x+(1)x+(1)$

Equating coefficients yields:

$\displaystyle -6c_4=1\,\therefore\,c_4=-\frac{1}{6}$

$\displaystyle 5c_4-7c_3=0\,\therefore\,c_3=-\frac{5}{36}$

$\displaystyle -10c_2=1\,\therefore\,c_2=-\frac{1}{10}$

$\displaystyle 3c_2-10c_1=1\,\therefore\,c_1=-\frac{13}{100}$

Thus, we have:

$\displaystyle y_p(x)=-\frac{13}{100}-\frac{1}{10}x-\frac{5}{36}e^x-\frac{1}{6}xe^x$

and so, the general solution to (1) is:

$\displaystyle y(x)=c_1e^{2x}+c_2e^{-5x}-\frac{13}{100}-\frac{1}{10}x-\frac{5}{36}e^x-\frac{1}{6}xe^x$

 

[Ackbach]

The answer is correct. And the method used is annihilator, so you seem to have done it the way you wanted. There aren't any efficiencies that I see available more than what you've already done.

 

[MarkFL]

Thank you, I just wanted to be sure I was doing this correctly and without any unnecessary steps.;)

 

[topsquark]

I've never heard of this method.

Coooooool! (Rock)

-Dan

 

[Ackbach]

I've never heard of this method.

Coooooool! (Rock)

-Dan

Check out Dennis Zill's A First Course in Differential Equations, Section 4.5. Annihilator and superposition are two methods classified as "Undetermined Coefficients". They work on ODE's which are linear with constant coefficients, and where the RHS "consists of finite sums and products of constants, polynomials, exponential functions $e^{ax}$, sines, and cosines."

 

[MarkFL]

I though the name Zill was familiar, Dennis G. Zill is the author of my old Calculus textbook.