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The general solution to


(where $a$ is a real-valued constant) is

$y=c_1e^{-2t}\sin t+c_2y_2+c_3y_3$

Find $a$, $y_2$, and $y_3$.

I thought that finding the characteristic equation would help. So I started as:


But it doesn't seem to really help with anything, so I'm not quite sure where to go from here. Can I make some assumptions based on the general solution?


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Since I get a sense this might be homework, I'll give a few hints.

  1. Based on the general solution, you should know one root of the characteristic equation.

  2. Complex roots of polynomials with real-valued coefficients come in pairs.

  3. Based on that last term of the characteristic polynomial, the product of the 3 roots is $-5$.

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Can you explain why the product is -5 rather than just 5? – Bob John Oct 8 '12 at 5:23
@BobJohn If the characteristic polynomial is $(x-r_1)(x-r_2)(x-r_3)=0$, the constant term is going to be $-r_1r_2r_3$ – Mike Oct 8 '12 at 7:19

You could just plug $e^{-2t}\sin t$ into the differential equation to get $a$.

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$$ y=c_1e^{-2t}\sin t+c_2y_2(t)+c_3y_3(t)\,. $$

That means, you've already had one of the fundamental solutions of the homogeneous ode. So, you can exploit it to find $a$ as a first step. Substituting $ y_1(x)=e^{-2 t}\sin(t) $ in the ode and simplifying you will find that $a=4$. Then the ode becomes

$$ y'''+5y''+9y'+5y=0 \Rightarrow (D^3+5D+9D^2+5)y= Ay=0 \,,$$


$$ A := D^3+5D+9D^2+5\,. $$

Now, the task is how to find the other two solutions. Recalling the annihilator method we used in the other problem, we have,

$$ y_1(x) = e^{-2 t}\sin(t) = \frac{1}{2i}e^{(-2+i)t} - \frac{1}{2i}e^{(-2-i)t} $$

Applying the annihilator $ (D-(-2+i))(D-(-2-i)) $ to the above equation gives

$$ (D-(-2+i))(D-(-2-i))y_1(t)= 0 \Rightarrow (D^2+4D+5)y_1= By_1 =0 \,, $$


$$ B = D^2+4D+5 $$

Now, if you divide $A$ by $B$ (division of polynomials), you get the other root of A. If you do that, you will get

$$A = (D+1)B= (D+1)(D-(-2+i))(D-(-2-i)) \,.$$

So, we were able to factor our operator. In fact, now you have the three roots you are looking for

$$ r_1 = -1 \,,\, r_2 = -2+i\,,\, r_3 = -2-i \,,$$

and the their corresponding solutions

$$ \left\{ y_1(t) = e^{-t} \,,\, y_2(t) = e^{(-2+i)t} \,,\, y_2(t) = e^{(-2-i)t}\right\}\,.$$

The general solution is given by

$$ y(x) = b_1 e^{-t} + b_2 e^{(-2+i)t} + b_3 e^{(-2-i)t} \,. $$

Now, manipulating the above general solution, exploiting the identity $e^{ix}=\cos(x)+i\sin(x)$, and then comparing with the general solution you were given

$$ y=c_1e^{-2t}\sin t+c_2y_2(t)+c_3y_3(t)\,, $$

you should be able to find $y_2(t)$ and $y_3(t)$. Solution is

$$ y_2(t) =e^{-t}\,,\, y_3(t)= e^{-2t}\cos(t) \,.$$

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