2021-12-29

Hyperbolic functions are useful in solving differential equations. Show that the functions , where A, B, and k are constants, satisfy the equation $yx\right)-{k}^{2}y\left(x\right)=0$

Bukvald5z

Step 1
Given,

Where A,B and k are constants
Our Aim
To satisfy Equation
$yx\right)-{k}^{2}y\left(x\right)=0$
Step 2
For $y=A\mathrm{sin}h\left(kx\right)$

Substitute values in

$⇒yx\right)-{k}^{2}y\left(x\right)=0$
Therefore, $y=A\mathrm{sin}h\left(kx\right)$ satisfies the equation
Step 3
For $y=B\mathrm{cos}h\left(kx\right)$
${y}^{\prime }=Bk\mathrm{sin}h\left(kx\right)$
$yB{k}^{2}\mathrm{cos}h\left(kx\right)$
Substituting values in
$yx\right)-{k}^{2}y\left(x\right)=B{k}^{2}\mathrm{cos}h\left(kx\right)-{k}^{2}B\mathrm{cos}h\left(kx\right)$
$⇒yx\right)-{k}^{2}y\left(x\right)=0$
Therefore, $y=B\mathrm{cos}h\left(kx\right)$ satiisfies the equation

Bukvald5z

$y=A\mathrm{sin}h\left(kx\right)$
${y}^{\prime }=A\cdot \mathrm{cos}h\left(kx\right)\cdot k=Ak\cdot \text{cosh}\left(kx\right)$
$y{}^{″}=Ak\cdot \mathrm{sin}h\left(kx\right)\cdot k$
$y{}^{″}=A{k}^{2}\cdot \mathrm{sin}h\left(kx\right)$
Diff equation is $y{}^{″}-{k}^{2}\cdot y=0$ I believe that this is what it has to be, it must not be $k\cdot y\left(x\right)$
Plug in results :
$A{k}^{2}\mathrm{sin}h\left(kx\right)-{k}^{2}\cdot \mathrm{sin}h\left(kx\right)=0$
$A{k}^{2}\mathrm{sin}h\left(kx\right)-A{k}^{2}\mathrm{sin}h\left(kx\right)=0$
$0=0$
Hence proved!
$y=B\mathrm{cos}h\left(kx\right)$
${y}^{\prime }=Bk\cdot \mathrm{sin}h\left(kx\right)$
$y{}^{″}=B{k}^{2}\cdot \mathrm{cos}h\left(kx\right)$
$y{}^{″}-{k}^{2}\cdot y=0$
$B{k}^{2}\cdot \mathrm{cos}h\left(kx\right)-{k}^{2}\cdot B\mathrm{cos}h\left(kx\right)=0$
$B{k}^{2}\cdot \mathrm{cos}h\left(kx\right)-B{k}^{2}\cdot \mathrm{cos}h\left(kx\right)=0$
$0=0$
Hence proved that $y=A\mathrm{sin}h\left(kx\right)$ and $B\mathrm{cos}h\left(kx\right)$ are solutions

karton