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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.

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

  1. 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

        \[ y'' - y = 0. \]

  2. 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

        \[ y'' - 4y' + 4y = 0. \]

  3. 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

        \[ y'' + y' + \frac{5}{4}y = 0. \]

  4. 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

        \[ y'' + 4y = 0. \]

  5. 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

        \[ y'' - y = 0. \]

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