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.

[tex]y=a\cosh cx+b\sinh cx[/tex]
[tex]y'=ac\sinh cx+bc\cosh cx[/tex]
[tex]y''=ac^2\cosh cx+bc^2\sinh cx[/tex]

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

[tex]y(0)=3[/tex]
[tex]\implies a\cosh0+b\sinh0=a=3[/tex]

[tex]y'(0)=3[/tex]
[tex]\implies ac\sinh0+bc\cosh0=bc=3[/tex]

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

We immediately find that [tex]c=\pm2[/tex] are possible choices, which makes [tex]b=\pm\dfrac32[/tex], respectively.

If [tex]c\neq\pm2[/tex], we can eliminate the factor of [tex]c^2-4[/tex] to get

[tex]\cosh cx+\dfrac1c\sinh cx=0\implies\tanh cx=-c[/tex]

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

That leaves us with [tex]a=3[/tex], [tex]b=\dfrac32[/tex], and [tex]c=2[/tex].