Question

Determine a, b, and c so that y=acosh(cx)+bsinh(cx)satisfies the conditions y′′−4y=0,y(0)=3,y′(0)=3. take c>0.

Answer 1

$y=a\cosh cx+b\sinh cx$
$y'=ac\sinh cx+bc\cosh cx$
$y''=ac^2\cosh cx+bc^2\sinh cx$

$y''-4y=0\iff ac^2\cosh cx+bc^2\sinh cx-4(a\cosh cx+b\sinh cx)=0$
$\implies (ac^2-4a)\cosh cx+(bc^2-4b)\sinh cx=0$

$y(0)=3$
$\implies a\cosh0+b\sinh0=a=3$

$y'(0)=3$
$\implies ac\sinh0+bc\cosh0=bc=3$

$(3c^2-12)\cosh cx+\left(3c-\dfrac{12}c\right)\sinh cx=0$
$(3c^2-12)\cosh cx+\dfrac1c(3c^2-12)\sinh cx=0$
$3(c^2-4)\left(\cosh cx+\dfrac1c\sinh cx\right)=0$

We immediately find that $c=\pm2$ are possible choices, which makes $b=\pm\dfrac32$, respectively.

If $c\neq\pm2$, we can eliminate the factor of $c^2-4$ to get

$\cosh cx+\dfrac1c\sinh cx=0\implies\tanh cx=-c$

Note that if $c=0$, we have equality, but this goes against our assumption that $c>0$. Note that $(\tanh x)'=\sech^2x>0$ for all $x$, which means $\tanh(cx)$ is a monotonically increasing function, and is bounded between -1 and 1. On the other hand, $-x$ is a monotonically decreasing function that is unbounded. From this you can gather that the two functions never intersect for $x>0$ (since $-x$ is always negative while $\tanh x$ is always positive), which means $c=0$ is the only solution to the equation above. However, this solution is actually extraneous, since the original equation contains a factor of $\dfrac1c$. So, in fact, the equation above has no solution for $c$.

That leaves us with $a=3$, $b=\dfrac32$, and $c=2$.