Kansas State University

Discussion

1. The solution to $(D-r)y = 0$ is $y = Ce^{rx}$.

2. The solution to $(D-r)^2 y = 0$ is $y = Cxe^{rx} + Ke^{rx}$.

3. The solution to $(D-r)^n y = 0$ is $y=C_{n-1}x^{n-1}e^{rx}+C_{n-2}x^{n-2}e^{rx}+\cdots+C_0e^{rx}$.

Paradigm

Step 1: Write in operator form (with $D$ for $d/dx$) $$(D^2+3D-10)y=0$$ Step 2: Factor and find roots.

$(D-2)(D+5)y=0$ so the roots are $2$, $-5$.

Step 3: Write the general solution. $$y(x)=Ce^{2x}+Ke^{-5x}$$

Step 1: $(D^2 + 4D + 3)y = 0 $

Step 2: $ (D+3)(D+1)y = 0 $ so the roots are $-3$, $-1$

Step 3: $ y = Ce^{-3x} + Ke^{-x} $

SECOND: Plug in the initial values and solve for the constants. $$ \begin{align} y(0) &= Ce^{-3\times0} + Ke^{-1\times0} = C + K \buildrel{\text{set}}\over{=} 1 \\ y'(x) &= -3Ce^{-3x} - Ke^{-x} \\ y'(0) &= -3Ce^{-3\times0} - Ke^{-1\times0} = -3C - K \buildrel{\text{set}}\over{=} 1 \end{align} $$ Taking 3 times the first equation and adding it to the second we find $2K=4$ so $K=2$ and $C=-1$. This gives us the solution $$y=-e^{-3x}+2e^{-x}$$

Step 2:$ (D+5)^2 y = 0$, double root of $-5$

Step 3: $y = Cxe^{-5x} + Ke^{-5x}$

Now we consider the equation $y''+y=0$. If we try to solve this equation using our new techniques, we obtain the operator $D^2+1$ which has roots $\pm i$, so the general solution is $c_1e^{ix}+c_2e^{-ix}$. This is in fact the correct general complex solution, provided we allow $c_1$ and $c_2$ to be arbitrary complex constants. But since we started with a real problem we probably want a real answer. One way to deal with that is to obtain the general complex solution as above and find all the complex solutions which are actually real solutions. The algebra involved in that is rather messy though. Fortunately, there is a quicker way that will give us two linearly independent real solutions from a single complex solution.

Theorem: If $e^{(\alpha+\beta i)x}$ is a solution of the equation $ay''+by'+cy=0$ where $a$, $b$ and $c$ are all real, then $e^{\alpha x}\cos(\beta x)$ and $e^{\alpha x}\sin(\beta x)$ are two linearly independent solutions of $ay''+by'+cy=0$.

Proof: We are given that $$ a(e^{(\alpha+\beta i)x})''+ b(e^{(\alpha+\beta i)x})'+c(e^{(\alpha+\beta i)x})=0 $$ Since $\Re[y'(x)]=(\Re[y(x)])'$ and $\Im[y'(x)]=(\Im[y(x)])'$, it follows that $$ \eqalign { a(\Re[e^{(\alpha+\beta i)x}])''+ b(\Re[e^{\alpha+\beta i)x}])'+c(\Re[e^{(\alpha+\beta i)x}])&=0 \cr a(\Im[e^{(\alpha+\beta i)x}])''+ b(\Im[e^{\alpha+\beta i)x}])'+c(\Im[e^{(\alpha+\beta i)x}])&=0 \cr} $$ So $e^{\alpha x}\cos(\beta x)$ and $e^{\alpha x}\sin(\beta x)$ are both real solutions to the given equation. Linear independence can be quickly checked using the Wronskian.

Using this theorem we can now handle complex roots efficiently. One point that often bothers students at this stage is that we only need one complex solution to find the complete general solution. What happens to the other complex solution? It is a theorem that complex roots to a real polynomial always come in complex conjugate pairs, $\alpha\pm\beta i$. So since the operator has real coefficients, if $\alpha+\beta i$ is one root of the polynomial equation, $\alpha-\beta i$ must be another root of the equation, and $\alpha-\beta i$ will give rise to the same pair of real roots. So we really get our pair of real roots from a pair of complex roots.

Another point to be noted is that we can write the pair of real solutions we get from our complex conjugate pair of roots $\alpha\pm\beta i$ as either $C_1e^{\alpha x}\cos(\beta x) + C_2e^{\alpha x}\sin(\beta x)$ or as $Ae^{\alpha x}\cos(\beta x-\phi)$ (with arbitrary constants $A$ and $\phi$) using the techniques of section 3. The first form is usually the easiest for solving initial value problems while the second form is usually the easiest for sketching solutions. And, of course, in initial value problems where you have to sketch solutions, you may want to solve first in one form to solve the initial value problem and then switch to the second form to draw your sketch.

EXAMPLE 3: $$ y'' + 2y' + 2y = 0$$ Step 1: Write the equation in operator form $$(D^2 + 2D + 2)y = 0$$ Step 2: Find the roots of the equation

roots are $-1\pm i$

Step 3: The general solution is $Ae^{-x}\cos(x-\phi)$ (or $C_1e^{-x}\cos(x)+C_2e^{-x}\sin(x)$ in the alternate form).

EXAMPLE 4: $$y''+y=0,\qquad y(0)=1,\quad y'(0)=1$$ FIRST: Find the general solution.

Step 1: $(D^2+1)y=0$

Step 2: roots are $\pm i$

Step 3: $y(x)=C\cos(x)+K\sin(x)$ (the easiest form for initial value problems).

SECOND: Plug in the initial values and solve for the constants $$ \begin{align} y(0)&=C \buildrel{\text{set}}\over{=} 1 \\ y'(x)&=-C\sin(x)+K\cos(x) \\ y'(0)&=K \buildrel{\text{set}}\over{=} 1 \end{align} $$ So the solution to the initial value problem is $y(x)=\cos(x)+\sin(x)$.

EXAMPLE 5: $$y'''+y=0$$ Step 1: $(D^3+1)y=0$

Step 2: $D^3+1=(D+1)(D^2-D+1)$ so the roots are $-1$ and $1/2\pm(\sqrt{3}/2)i$.

Step 3: The solution corresponding to the root $-1$ is $e^{-x}$. The solutions corresponding to the complex conjugate pair $1/2\pm\sqrt{3}/2$ are $Ae^{x/2}\cos(\sqrt{3}x/2-\phi)$. So the general solution is $$y(x)=C_1e^{-x}+Ae^{x/2}\cos(\sqrt{3}x/2-\phi).$$

EXAMPLE 6: $$y''''+8y'''+27y''+38y'+26y=0$$ Step 1: $(D^4+8D^3+27D^2+38D+26)y=0$

Step 2: $D^4+8D^3+27D^2+38D+26=(D^2+2D+2)(D^2+6D+13)=0$ so the roots are $-1\pm i$ and $-3\pm 2i$.

Step 3: $y=Ae^{-x}\cos(x-\phi)+Be^{-3x}\cos(2x-\psi)$, or written in the alternate form $y=C_1e^{-x}\cos(x)+C_2e^{-x}\sin(x)+C_3e^{-3x}\cos(2x)+ C_4e^{-3x}\sin(2x)$.

EXAMPLE 7: $$y''''+4y'''+24y''+40y'+100y=0$$ Step 1: $(D^4+4D^3+24D^2+40D+100)y=0$

Step 2: $(D^2+2D+10)^2=0$ so we have double roots of $-1\pm 3i$.

Step 3: We handled double real roots by using both the usual solution and the solution multiplied by $x$. We do exactly the same thing with double complex roots. So the solution is $y=Ae^{-x}\cos(3x-\phi)+Bxe^{-x}\cos(3x-\psi)$ or in the alternate form $y=C_1e^{-x}\cos(3x)+C_2e^{-x}\sin(3x)+C_3xe^{-x}\cos(3x)+C_4xe^{-x}\sin(3x)$.

You can generate additional examples of second-order constant-coefficient linear homogeneous initial value problems here.