We consider second order linear differential equations of the form
$$y'' = f(t, y, y') = g(t) - p(t)y' - q(t)y$$or equivalently
$$y'' + p(t)y' + q(t)y = g(t),$$where $p$, $q$ and $g$ are continuous functions of $t$.
Example. Solve $y'' + 5y' + 6y = 0.$
The main idea is to try
$$y(t) = e^{rt},$$for some constant $r$. Then $y'(t) = re^{rt} = ry$ and $y''(t) = r^2 e^{rt} = r^2 y$. Substituting this $y$ into the DE yields
$$(r^2 + 5r + 6)\,e^{rt} = 0.$$Since $e^{rt} \neq 0$, we must have
$$r^2 + 5r + 6 = 0.$$Thus $r = -2$ and $r = -3$. Hence $y_1 = e^{-2t}$ and $y_2 = e^{-3t}$ are both solutions to the DE above. We can easily verify (as your exercise) that any linear combination
where $c_1$ and $c_2$ are arbitrary constants, is also a solution to the DE.
Given constants $a$, $b$, $c$ and the homogeneous DE
$$ay'' + by' + cy = 0. \tag{1}$$Let $y = e^{rt}$. Then the quadratic equation
$$ar^2 + br + c = 0$$is called the characteristic equation and its roots $r_1$ and $r_2$ are called characteristic roots of the DE.
This immediately implies:
Suppose $r_1$ and $r_2$ where
$$r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$are characteristic roots of $ar^2 + br + c = 0$. Then
$$y_1 = e^{r_1 t}, \qquad y_2 = e^{r_2 t}$$are both solutions to the DE (1).
We note that the characteristic roots $r_1$, $r_2$ could be complex numbers. This would lead to important consequences just as when the roots are real. There is also a possibility of having a real double root $r_1 = r_2$. We shall discuss these two important issues in due course.
Suppose $y_1$ and $y_2$ are solutions to
$$ay'' + by' + cy = 0. \tag{2}$$Then $c_1 y_1 + c_2 y_2$, where $c_1$, $c_2$ are arbitrary (including complex) constants, is also a solution to the DE (2).
Then the DE (2) can be simply rewritten as $L(y) = 0$. Suppose that $y_1$, $y_2$ are both solutions to (2). Let $c_1$, $c_2$ be arbitrary constants. Then
\begin{align} L(c_1 y_1 + c_2 y_2) &= \Big(a\frac{d^2}{dx^2} + b\frac{d}{dx} + c\Big)(c_1 y_1 + c_2 y_2) \\[6pt] &= c_1(ay_1'' + by_1' + cy_1) + c_2(ay_2'' + by_2' + cy_2) \\[6pt] &= c_1 L(y_1) + c_2 L(y_2) = 0 + 0 = 0. \end{align}Hence the $c_1 y_1 + c_2 y_2$ is also a solution to the (2).
Suppose we know that $y_1, y_2, \cdots, y_n$ are solutions to the DE. Then
$$c_1 y_1 + c_2 y_2 + \cdots + c_n y_n$$for arbitrary constants $c_1, c_2, \cdots, c_n$, is also a solution to the (2).
Let $a$, $b$, $c$ be given constants. Suppose the characteristic roots $r_1$, $r_2$ of the second order linear DE
$$ay'' + by' + cy = 0$$are distinct. Then
where $c_1$, $c_2$ are two arbitrary constants. Substitute $y$ into the DE yields
\begin{align} ay'' + by' + cy &= c_1(ar^2 + br + c)e^{r_1 t} + c_2(ar^2 + br + c)e^{r_2 t} \\[6pt] &= 0 + 0. \end{align}This shows that the $y(t)$ is a solution to the DE. Let $t = t_0$. Then
$$y_0 = y(t_0) = c_1\,e^{r_1 t_0} + c_2\,e^{r_2 t_0}. \tag{3}$$Differentiating $y$ and substituting $t = t_0$ yields
$$y_0' = y'(t_0) = c_1 r_1\,e^{r_1 t_0} + c_2 r_2\,e^{r_2 t_0}. \tag{4}$$For example, we could write
$$c_2 = y_0 e^{-r_2 t_0} - c_1 e^{(r_1 - r_2)t_0}$$from the (4). Substitute this $c_2$ into the equation (3) yielding, under the assumption $r_1 \neq r_2$,
$$c_1 = \Big(\frac{y_0' - r_2 y_0}{r_1 - r_2}\Big)e^{-r_1 t_0}$$after simplification. Substituting this $c_1$ back into the equation (4) and simplifying yields
$$c_2 = \Big(\frac{y_0 r_1 - y_0'}{r_1 - r_2}\Big)e^{-r_2 t_0}$$as asserted. This completes the proof.
Example (IVP). Solve $y'' + 5y' + 6y = 0$, $y(0) = 2$, $y'(0) = 3$.
We have already found the characteristic roots to be $-2$, $-3$ and hence two solutions $e^{-2t}$, $e^{-3t}$. According to the last Theorem, we know that
$$y(t) = c_1\,e^{-2t} + c_2\,e^{-3t}$$is also a solution to the DE. We substitute IC $y(0) = 2$ into this $y$. This yields
$$2 = c_1 + c_2.$$Differentiating $y$ and substitute IC $y'(0) = 3$ yields
$$3 = y'(0) = -2c_1 - 3c_2.$$Solving the above two equations in $c_1$, $c_2$ yields $c_1 = 9$ and $c_2 = -7$. Hence
One can easily verify that $y(0) = 2$, $y'(0) = 3$ hold.
Example. Solve $y'' - y = 0$, $y(0) = 2$, $y'(0) = \beta$. Find $\beta$ if $y \to 0$ as $t \to \infty$.
It is easy to check that the characteristic equation is
$$4r^2 - 1 = 0,$$and hence characteristic roots $r = \pm 1/2$. Thus we have
$$y(t) = c_1\,e^{t/2} + c_2\,e^{-t/2}.$$The IC $y(0) = 2$, $y'(0) = \beta$ yields
$$2 = y(0) = c_1 + c_2,$$and
$$\beta = y'(0) = c_1/2 - c_2/2.$$That is $2\beta = c_1 - c_2$. Solving these two equations in $c_1$, $c_2$ yields $c_1 = 1 + \beta$ and $c_2 = 1 - \beta$. Hence
$$y(t) = (1 + \beta)e^{t/2} + (1 - \beta)\,e^{-t/2}.$$If $y \to 0$ as $t \to \infty$, then $1 + \beta = 0$ or $\beta = -1$ and $y(t) = 2\,e^{-t/2}$.
Solve the following IVP
Suppose that the DE
$$ay'' + by' + cy = 0,$$has characteristic equation
$$ar^2 + br + c = 0$$that admits two complex (conjugate) roots
$$r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} := \lambda \pm i\mu$$(i.e., $b^2 - 4ac < 0$). Then any solution $y$ can be written in the form
$$y(t) = d_1 e^{\lambda t}\cos\mu t + d_2 e^{\lambda t}\sin\mu t.$$We see that $\overline{r_1} = r_2$, that is, the two complex roots must be conjugate to each other, and hence they must be distinct.
We now note the superposition principle asserts that if $e^{r_1 t}$, $e^{r_2 t}$ are two solutions to the DE, then so are the functions
$$\frac{1}{2}(e^{r_1 t} + e^{r_2 t}), \qquad \frac{1}{2i}(e^{r_1 t} - e^{r_2 t}).$$But
$$\frac{1}{2}(e^{r_1 t} + e^{r_2 t}) = e^{\lambda t}\frac{1}{2}(e^{i\mu t} + e^{-i\mu t}) = e^{\lambda t}\cos\mu t,$$and
$$\frac{1}{2i}(e^{r_1 t} - e^{r_2 t}) = e^{\lambda t}\frac{1}{2i}(e^{i\mu t} - e^{-i\mu t}) = e^{\lambda t}\sin\mu t.$$We deduce that the $\{e^{\lambda x}\cos\mu x,\; e^{\lambda x}\sin\mu x\}$ is another pair of (linearly independent, i.e., "different") solutions to the DE. In particular, the general solution to the DE can be expressed in the form
$$y(t) = d_1 e^{\lambda t}\cos\mu t + d_2 e^{\lambda t}\sin\mu t$$where $d_1$ and $d_2$ can be any constants.
Another way is to start from the general solution
\begin{align} y &= c_1 e^{r_1 t} + c_2 e^{r_2 t} = c_1 e^{(\lambda + i\mu)t} + c_2 e^{(\lambda - i\mu)t} \\[6pt] &= c_1 e^{\lambda t}(\cos\mu t + i\sin\mu t) + c_2 e^{\lambda t}(\cos\mu t - i\sin\mu t) \\[6pt] &= (c_1 + c_2)e^{\lambda t}\cos\mu t + i(c_1 - c_2)e^{\lambda t}\sin\mu t \\[6pt] &= C_1 e^{\lambda t}\cos\mu t + C_2 e^{\lambda t}\sin\mu t \end{align}where $C_1 = c_1 + c_2$ and $C_2 = i(c_1 - c_2)$. This discussion shows that both $e^{\lambda}\cos\mu$ and $e^{\lambda}\sin\mu$ are also solution to the DE for complex characteristic roots $r_1$ and $r_2$ defined above.
Example. Solve $y'' + y' + y = 0$.
Clearly the characteristic equation is
$$r^2 + r + 1 = 0$$so that the characteristic roots are
$$r_{1,2} = -\frac{1}{2} \pm \frac{\sqrt{1 - 4}}{2} = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}.$$So the general solution is
$$y = d_1 e^{-\frac{1}{2}t}\cos\frac{\sqrt{3}t}{2} + d_2 e^{-\frac{1}{2}t}\sin\frac{\sqrt{3}t}{2}.$$Notice that
$$y(t) = e^{-\frac{1}{2}t}\Big(d_1\cos\frac{\sqrt{3}t}{2} + d_2\sin\frac{\sqrt{3}t}{2}\Big) \to 0$$as $t \to \infty$. This result is independent of $d_1$, $d_2$.
Example. Solve $y'' + 9y = 0$.
The characteristic equation of the DE is $r^2 + 9 = 0$. Hence $r = \pm 3i$. So the general solution is given by
\begin{align} y &= e^{0}(d_1\cos 3t + d_2\sin 3t) \\[6pt] &= d_1\cos 3t + d_2\sin 3t \end{align}We note that this solution is bounded (oscillates between fixed values).
Example (IVP). Solve the IVP $16y'' - 8y' + 145y = 0$, $y(0) = -2$, $y'(0) = 1$.
We need to solve
$$16r^2 - 8r + 145 = 0.$$So
$$r = \frac{8 \pm \sqrt{64 - 4(16)(145)}}{32} = \frac{1}{4} \pm 3i.$$Thus
$$y = e^{t/4}(c_1\cos 3t + c_2\sin 3t).$$The IC gives
$$-2 = y(0) = c_1\cos 0 + 0 = c_1.$$Thus $c_1 = -2$. Differentiating $y$ yields
$$y' = \frac{1}{4}e^{t/4}(c_1\cos 3t + c_2\sin 3t) + e^{t/4}(-3c_1\sin 3t + 3c_2\cos 3t).$$Substitute $t = 0$ yields
$$1 = y'(0) = \frac{1}{4}c_1 + 3c_2.$$Hence $c_2 = 1/2$ and
We see that $y(t) \to \infty$ as $t \to \infty$ (growing oscillation).
Exercise (Ex. 3.4, Q. 7). Solve $y'' - 2y' + 2y = 0$.
Thus the characteristic roots are
$$r_{1,2} = \frac{2 \pm \sqrt{4 - 4(1)(2)}}{2} = 1 \pm i.$$Thus
$$y(t) = e^t(c_1\cos t + c_2\sin t).$$Solve for the general solution of:
Solve for the following initial value problems and describe its behaviour for increasing $t$.
Let us recall an trigonometric identity, namely the
$$\cos(t + \phi) = \cos t \cos\phi + \cos t \cos\phi.$$Then for any given constants $A$ and $B$, we can write
\begin{align} A\cos t + B\sin t &= \sqrt{A^2 + B^2}\Big[\frac{A}{\sqrt{A^2 + B^2}}\cos t + \frac{B}{\sqrt{A^2 + B^2}}\sin t\Big] \\[6pt] &= \cos t \cos\phi + \sin t \sin\phi \\[6pt] &= \sqrt{A^2 + B^2}\cos(t + \phi) \end{align}where we have identified
$$\cos\phi = \frac{A}{\sqrt{A^2 + B^2}}, \quad \sin\phi = \frac{B}{\sqrt{A^2 + B^2}}, \quad \tan\phi = \frac{B}{A}.$$Suppose that the DE
$$ay'' + by' + cy = 0,$$has characteristic equation $ar^2 + br + c = 0$ which admits two complex (conjugate) roots
$$r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} := \lambda \pm i\mu$$(i.e., $b^2 - 4ac < 0$). Then the general solution $y$ can be written in the form
$$y(t) = Ce^{\lambda t}\cos(\mu t + \phi),$$for some constants $C$ and $\phi$.
where $\phi = \arctan(d_2/d_1)$.
The constants $d_1$, $d_2$ (i.e., $\phi$) are determined by initial conditions of the given problem concerned.
Example (revisited). Let us reconsider the IVP
$$16y'' - 8y' + 145y = 0, \quad y(0) = -2,\; y'(0) = 1$$that we studied earlier in this section. We have found that the characteristic roots are $\frac{1}{4} \pm 3i$. Hence the initial conditions $y(0) = -2$, $y'(0) = 1$ implies that the solution is
$$y(t) = e^{t/4}\Big(-2\cos 3t + \frac{1}{2}\sin 3t\Big).$$We have $d_1 = 1/2$, $d_2 = -2$ so that $C = \sqrt{d_1^2 + d_2^2} = \sqrt{17}/2$. Thus
$$\phi = \arctan(-2/\tfrac{1}{2}) = \arctan(-4) \approx -1.33 \approx -76°.$$So
$$y(t) \approx \frac{\sqrt{17}}{2}e^{t/4}\sin(3t - 1.33) = \frac{\sqrt{17}}{2}e^{t/4}\sin 3(t - 1.33/3).$$Determine $\omega_0$, $R$ and $\delta$ of the following trigonometric expressions when written into the form $u = R\cos(\omega_0 t - \delta)$.
The characteristic equation is $ar^2 + br + c = 0$ with discriminant $\Delta = b^2 - 4ac$.
Solve the differential equation:
$$y'' + 3y' + 2y = 0$$What type of DE is this and what should we try?
What's the first step?
After getting the characteristic equation, what's next?
Once we have distinct real roots $r_1 \neq r_2$, the general solution is?
For $y'' + 3y' + 2y = 0$, the characteristic equation is:
Factor $r^2 + 3r + 2 = 0$. The roots are:
The general solution is:
Solve the differential equation:
$$y'' + 2y' + 5y = 0$$The characteristic equation is $r^2 + 2r + 5 = 0$. What type of roots does it have?
For complex roots $r = \lambda \pm i\mu$, what form does the general solution take?
In $r = \lambda \pm i\mu$, what do $\lambda$ and $\mu$ represent from the quadratic formula?
For $r^2 + 2r + 5 = 0$ (where $a=1$, $b=2$, $c=5$), compute $\lambda = -b/(2a)$:
Compute $\mu = \sqrt{4ac - b^2}/(2a) = \sqrt{4(1)(5) - 4}/2$:
With $\lambda = -1$ and $\mu = 2$, the general solution is:
Solve the initial value problem:
$$y'' + 5y' + 6y = 0, \quad y(0) = 2, \quad y'(0) = 3$$The characteristic equation $r^2 + 5r + 6 = 0$ factors as $(r+2)(r+3) = 0$. The general solution is:
From $y(0) = 2$, substituting $t = 0$ into $y = c_1 e^{-2t} + c_2 e^{-3t}$ gives:
We need $y'$. Differentiating $y = c_1 e^{-2t} + c_2 e^{-3t}$:
From $y'(0) = 3$, substituting $t = 0$ gives:
Solve the system: $c_1 + c_2 = 2$ and $-2c_1 - 3c_2 = 3$. Find $c_1$:
Find $c_2$:
The solution to the IVP is:
— End of Second Order Homogeneous Equations Notes —