# The differential equation $y''-q(x)y=0.$

Let $$y''-q(x)y=0$$be a differential equation with initial conditions on $0\leq x<\infty,$ as $y(0)=1,y'(0)=1$ where $q(x)$ is a positive monotonically increasing continuous function. Then which of the following are true?

1. $y(x)\rightarrow\infty$ as $x\rightarrow\infty$.

2. $y'(x)\rightarrow\infty$ as $x\rightarrow\infty$.

3. $y(x)$ has finitely many zeros in $[0,\infty)$.

4. $y(x)$ has infinitely many zeros in $[0,\infty)$.

Please don't mind, actually I am new in differential equation. I only know that by Picard's theorem the above differential equation has the unique solution, but I don't know what is the solution as I tried by direct hit and trial method. According to me the solution of above differential equation will be some thing in exponential form, so according to me its answer will be $a$, $b$, $c$. But I don't know the exact method. Please help me to solve the above problem. Thanks in advance.

• no its right... – neelkanth Mar 29 '16 at 15:17

We can rewrite the ODE as $y''=q(x)y$. When $x=0$, both $y$ and $y'$ are positive. The ODE tells us that $y''$ is also positive. Therefore, the function is increasing, and its derivative is also increasing. As the function gets bigger, $q(x)$ also gets bigger. This leads to $y''$ also getting bigger, which means that $y'$ gets bigger too! There is nothing to stop the growth process; the solution just grows without bound forever. This should give you enough information to determine which of those options are correct.
• how can we say that $y$ and $y^{'}$ are increasing on $[0,\infty)$? – neelkanth Mar 29 '16 at 15:23
• Since $y(0)=1$, $y''(0)>0$. This means that $y'$ is increasing, at least initially. But $y'$ increasing means $y$ is increasing, which means that $y''$ is increasing which goes back to $y'$ increasing. It's like a feedback loop. – Alex S Mar 29 '16 at 15:26