Find the extremal of $\int_a^b (y’^2+2yy’-16y^2)\, dx$

Question

The task is to find the extremals of:
$$\int_a^b (y^{\prime2}+2yy^\prime-16y^2)\, dx$$
I used the Euler-Lagrange equation and I got:
$$-2y^{\prime\prime} – 32y = 0$$
I solved the differential equation
$$y = c_2 \sin(4x) + c_1 \cos(4x)$$

But according to my textbook the solution should be $y = c_1\sin(4x-c_2)$. Can anyone explain please, where my mistake is?

Thanks.

Answer 1

You can expand the textbook solution to get

$$\begin{equation}\begin{aligned} y & = c_1\sin(4x - c_2) \\ & = c_1(\sin(4x)\cos(c_2) - \cos(4x)\sin(c2)) \\ & = (c_1\cos(c_2))\sin(4x) + (-c_1\sin(c2))\cos(4x) \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Note you can choose $c_1$ and $c_2$ values to allow any set of $2$ coefficients for the $\sin(4x)$ and $\cos(4x)$ terms, so this is basically equivalent to your solution.

To see why you have the full range of possible values, the second coefficient divided by the first gives

$$\frac{-c_1\sin(c2)}{c_1\cos(c2)} = -\tan(c_2) \tag{2}\label{eq2A}$$

Since the range of $\tan()$ is all real numbers, this means you can use the corresponding ratio of your two coefficients (call them $c_2'$ and $c_1'$ to help avoid confusion) to get

$$\frac{c_2'}{c_1'} = -\tan(c_2) \tag{3}\label{eq3A}$$

This determines a value of $c_2$ in the range of $0$ to $2\pi$, and with this, if $\cos(c_2) \neq 0$, you can then get $c_1$ from

$$c_1 = \frac{c_2'}{\cos(c_2)} \tag{4}\label{eq4A}$$

else, you can get it instead from

$$c_1 = \frac{-c_1'}{\sin(c_2)} \tag{5}\label{eq5A}$$

I'm leaving out certain details such as with $\cos(c_2) = 0$, i.e., the $\cos(4x)$ term having a coefficient of $0$, but these don't affect the generality of the textbook answer compared to yours.

For a more graphical/visual way to see this, consider the equation of a circle of radius $|c_1|$ centered at the origin is

$$x^2 + y^2 = c_1^2 \tag{6}\label{eq6A}$$

You have that $(c_1\cos(c_2), -c_1\sin(c_2))$, as $c_2$ goes from $0$ to $2\pi$, are the set of all of the points on this circle. As $|c_1|$ goes from $0$ to $\infty$, the set of points on the circle sweeps the entire plane, i.e., all points.