- Thread starter
- #1

- Thread starter oasi
- Start date

- Thread starter
- #1

For example, take $ay' + by = 0$. Solving for y' yields

$$

y' = -\frac{b}{a}y = ky

$$

where k = -b/a.

The only nontrivial function whose derivative is a constant multiple of itself is an exponential function.

- Jan 26, 2012

- 183

But $y = \sin x$ could work. For example,

$y'' + y = 0$

has as one solution $y = \sin x$.

$y'' + y = 0$

has as one solution $y = \sin x$.

As long as the boundaries aren't $y'(0) = 0$ and $y'\left(\frac{\pi}{2}\right) = 0$ then y = 0.But $y = \sin x$ could work. For example,

$y'' + y = 0$

has as one solution $y = \sin x$.

But $y = A\sin x + B\cos x = e^0\left(A\sin x + B\cos x\right)$ is also a solution of the non boundary value problem.