# simple Initial Value problem

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??

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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

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 $y(x) = c_1 e^x + c_2 e^{-2x} + c_3 e^{2x}$. $$\lim_{x \to \infty} y(x) = \lim_{x \to \infty} \left(c_1 e^x + c_2 e^{-2x} + c_3 e^{2x}\right) = \lim_{x \to \infty} \left(c_1 e^x + c_3 e^{2x}\right)= \text{sign} (c_3) \times \infty$$