Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

DE: $y^{(4)} - 4y''' + 3y'' + 4y' - 4 = 0$, $y(0) = 1, y(\infty) = 0$

obvoiusly, solution is $y(x) = c_1e^x + c_2e^{-2x} +c_3e^{2x} + c_4xe^{2x}$

I dont understand why $y(\infty) = 0$ implies $c_1 = c_2 = c_3 = 0$

Why is this true??

share|cite|improve this question
I assume you mean $c_1 = c_3 = c_4 = 0$. In fact, $c_2 = 1$, from the other initial condition. – user17762 Oct 15 '12 at 5:52
up vote 2 down vote accepted

If $c_1$ or $c_3$ or $c_4$ were non-zero, then $\lim_{x \to \infty} y(x)$ would be $\pm \infty$ or not defined.

For instance, consider the case when $c_3 = c_4 = 0$. Then the solution is $y(x) = c_1 e^x + c_2 e^{-2x}$. $$\lim_{x \to \infty} y(x) = \lim_{x \to \infty} \left(c_1 e^x + c_2 e^{-2x} \right) = \lim_{x \to \infty} c_1 e^x + 0 = \text{sign} (c_1) \times \infty$$

Next consider the case when $c_4 = 0$. Then the solution is $y(x) = c_1 e^x + c_2 e^{-2x} + c_3 e^{2x}$. $$\lim_{x \to \infty} y(x) = \lim_{x \to \infty} \left(c_1 e^x + c_2 e^{-2x} + c_3 e^{2x}\right) = \lim_{x \to \infty} \left(c_1 e^x + c_3 e^{2x}\right)= \text{sign} (c_3) \times \infty$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.