Determine functions u and v satisfying given differential equations

The graph of a solution $u$ of the differential equation

$y'' - 4y' + 29y = 0$

intersects the graph of a solution $v$ of the differential equation

$y'' + 4y' + 13y = 0$

at the origin. Furthermore, the two curves have equal slopes at the origin. Determine formulas for the functions $u$ and $v$ if

$u'\left( \frac{1}{2} \pi \right) = 1.$

First, we compute the general solutions of the second-order differential equation

$y'' - 4y' + 29y = 0.$

This is an equation of the form

$y'' + ay' + by = 0 \qquad \text{with} \qquad a = -4, \ \ b = 29.$

From this we compute $d = a^2 - 4b = -100$ , so $d < 0$ and $k = \frac{1}{2} \sqrt{-d} = 5$ . Using Theorem 8.7 (pages 326-327 of Apostol) we have

$y = e^{2x} (c_1 \cos (5x) + c_2 \sin (5x)).$

Next, we compute the general solutions of the second-order differential equation

$y'' + 4y' + 13y = 0.$

This is again an equation of the form

$y'' + ay' + by = 0 \qquad \text{with} \qquad a = 4, \ \ b = 13.$

From this we compute $d = a^2 -4b = -36$ so $d < 0$ and $k = \frac{1}{2} \sqrt{-d} = 3$ . Again, applying Theorem 8.7 (pages 326-327 of Apostol) we have

$y = e^{-2x} (c_1 \cos (3x) + c_2 \sin (3x)).$

Now, let

$u = e^{2x}(a_1 \cos (5x) + a_2 \sin (5x)), \qquad v = e^{-2x} (b_1 \cos (3x) + b_2 \sin (3x)).$

We know that the solutions we want intersect at the origin, hence we must have $u(0) = v(0) = 0$ . So,

$\begin{align*} u(0) &= 0 &\implies && a_1 &= 0, \\ v(0) &= 0 & \implies && b_1 &= 0. \end{align*}$

Plugging these values for the constants $a_1$ and $b_1$ into our expressions for $u$ and $v$ we compute the first derivatives,

$\begin{align*} u' &= a_2 (2e^{2x} \sin (5x) + 5e^{2x} \cos (5x)) \\ v' &= b_2 (-2e^{-2x} \sin (3x) + 3e^{-2x} \cos (3x)). \end{align*}$

Next, we are given that the particular solutions $u$ and $v$ we want have equal slopes at the origin. This means $u'(0) = v'(0)$ . Hence,

$u'(0) = v'(0) \quad \implies \quad 5a_2 = 3b_2 \quad \implies \quad b_2 = \frac{5}{3} a_2.$

Finally,

$u' \left( \frac{\pi}{2} \right) = 1 \quad \implies \quad 2a_2 e^{\pi} = 1 \quad \implies \quad a_2 = (2e^{pi})^{-1} \quad \implies \quad b_2 = \frac{5}{6e^{\pi}}.$

Thus,

$u(x) = \frac{1}{2} e^{2x-\pi} (\sin 5x), \qquad v(x) = \frac{5}{6} e^{-2x-\pi} (\sin 3x).$

Stumbling Robot

Determine functions u and v satisfying given differential equations

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