Find a linear differential equation satisfied by given equations

In each of the following cases find a second-order linear differential equation satisfied by the given $u_1$ and $u_2$ .

$u_1 (x) = e^x, \qquad u_2 (x) = e^{-x}$ ,
$u_1 (x) = e^{2x}, \qquad u_2 (x) = xe^{2x}$ ,
$u_1 (x) = e^{-\frac{x}{2}} \cos x, \qquad u_2(x) = e^{-\frac{x}{2}} \sin x$ .
$u_1 (x) = \sin (2x+1), \qquad u_2 (x) = \sin (2x+2)$ .
$u_1 (x) = \cosh x, \qquad u_2(x) = \sinh x$ .

We compute the first two derivatives of the given equations,
$\begin{align*} u_1 (x) &= e^x & \text{and} && u_2 (x) &= e^{-x} \\ u_1'(x) &= e^x & \text{and} && u_2' (x) &= -e^{-x} \\ u_1''(x) &= e^x & \text{and} && u_2''(x) &= e^{-x}. \end{align*}$

So, if $u_1(x)$ and $u_2(x)$ are solutions of a linear second order differential equation, $y'' + ay' + by = 0$ then we have

$e^x + ae^x + be^x = 0 \quad \implies \quad b = -1-a,$

and

$e^{-x} - ae^{-x} + be^{-x} = 0 \quad \implies \quad 1-a+b= 0.$

Together these imply

$a = 0, \quad b = -1.$

Therefore, $u_1(x), u_2(x)$ are solutions of

$y'' - y = 0.$
We compute the first two derivatives of the given equations,
$\begin{align*} u_1 (x) &= e^{2x} & \text{and} && u_2 (x) &= xe^{2x} \\ u_1'(x) &= 2e^{2x} & \text{and} && u_2' (x) &= e^{2x} + 2xe^{2x} \\ u_1''(x) &= 4e^{2x} & \text{and} && u_2''(x) &= 4e^{2x} + 4xe^{2x}. \end{align*}$

So, if $u_1(x)$ and $u_2(x)$ are solutions of a linear second order differential equation, $y'' + ay' + by = 0$ then we have

$4e^{2x} + 2ae^{2x} + be^{2x} = 0 \quad \implies \quad b = -2a -4,$

and

$\begin{align*} &&(4e^{2x} + 4xe^{2x}) + a (e^{2x} + 2xe^{2x}) + bxe^{2x} &= 0 \\ \implies && 4e^{2x} + 4xe^{2x} + ae^{2x} + 2xae^{2x} - 2axe^{2x} - 4xe^{2x} &= 0 \\ \implies && ae^{2x} + 4e^{2x} &= 0 \\ \implies && a = -4 \quad \implies \quad b = 4. \end{align*}$

Therefore, $u_1(x), u_2(x)$ are solutions of

$y'' - 4y' + 4y = 0.$
Rather than take derivatives of these two functions (which gets messy) we consider the function
$u = e^{-\frac{ax}{2}} (c_1 \cos (kx) + c_2 \sin (kx)).$

Then if we let $a = 1$ and $k = 1$ we have $\frac{1}{2} \sqrt{-d} = k = 1 \implies d = -4$ . Since $d = a^2 - 4b$ this gives us $b = \frac{5}{4}$ . By Theorem 8.7, then, $u$ is a solution of $y'' + y' + \frac{5}{4}y = 0$ for every choice of $c_1$ and $c_2$ . The functions $u_1$ and $u_2$ correspond to the choices $c_1 = 1, c_2 = 0$ and $c_1 = 0, c_2 = 1$ , respectively. Hence, $u_1(x)$ and $u_2(x)$ are solutions of the second order differential equation

$y'' + y' + \frac{5}{4}y = 0.$
Similar to our strategy in part (c), let
$u = e^{-\frac{ax}{2}} (c_1 \cos (kx) + c_2 \sin (kx)).$

Letting $a = 2$ and $k = 2$ we obtain $b = 4$ . So these are solutions of the equation $y'' + 4y = 0$ . Then, since

$u_1 = \sin (2x+1) = \cos 1 \sin (2x) + \sin 1 \cos (2x),\qquad u_2 = \sin 2 \cos (2x) + \cos 2 \sin (2x),$

we have $u_1$ and $u_2$ are solutions to the equation corresponding the choices $c_1 = \cos 1, c_2 = \sin 1$ and $c_1 \cos 2, c_2 = \sin 2$ , respectively. Therefore $u_1(x)$ and $u_2(x)$ are solutions of

$y'' + 4y = 0.$
First, using the definitions of the hyperbolic sine and cosine we have
$u_1(x) = \frac{1}{2}e^x + \frac{1}{2}e^{-x}, \qquad u_2(x) = \frac{1}{2}e^x - \frac{1}{2}e^{-x}.$

So if we let

$u = e^{-\frac{ax}{2}} (c_1 e^{kx} + c_2 e^{-kx})$

with $a = 0$ and $k = 1$ we have $b = -1$ . Therefore, $u$ is the set of solutions to $y'' - y = 0$ . Since $u_1(x)$ and $u_2(x)$ are the the solutions $u$ with $c_1 = c_2 =\frac{1}{2}$ and $c_1 = \frac{1}{2}, \ c_2 = -\frac{1}{2}$ , respectively we have $u_1(x)$ and $u_2(x)$ solutions

$y'' - y = 0.$

2 comments

August 8, 2022 at 9:21 am

Mohammad Azad says:

You can solve this problem in a much easier way by plugging some numbers, also Dsinh(x) = cosh(x) and Dcosh(x) = sinh(x) (This is why I love hyperbolic functions)

February 21, 2022 at 12:02 pm

Anonymous says:

for part (d), a=2
thank you for all you’ve done, hope all is well.

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Find a linear differential equation satisfied by given equations

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