# Thread: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

1. ## why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

How do i show this mathematically, preferably without using auxillary equations, that the solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax) 2. ## Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

I think you can use the substitution: $\displaystyle y=e^{tx}$.

3. ## Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

where to from there? I end up with:

$\displaystyle y\left(\left(\frac{\ln{y}}{x}\right)^2-9\right)=0\\\\y=e^{\pm3x}$

I'm not sure if this is right. There are no general constants !!!

4. ## Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

I would solve it like this:
Let $\displaystyle y=e^{tx}$
Substituting in the equation gives:
$\displaystyle t^2\cdot e^{tx}-a^2\cdot e^{tx}=0$
$\displaystyle t^2-a^2=0 \Leftrightarrow (t-a)(t+a)=0$
$\displaystyle t=a \ \mbox{or} \ t=-a$
So the general solution is:
$\displaystyle y=c_1\cdot e^{ax}+c_2\cdot e^{-ax}$
Where $\displaystyle c_1,c_2$ are constants (which can be chosen).

5. ## Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax) Originally Posted by Siron I would solve it like this:
Let $\displaystyle y=e^{tx}$
Substituting in the equation gives:
$\displaystyle t^2\cdot e^{tx}-a^2\cdot e^{tx}=0$
$\displaystyle t^2-a^2=0 \Leftrightarrow (t-a)(t+a)=0$
$\displaystyle t=a \ \mbox{or} \ t=-a$
So the general solution is:
$\displaystyle y=c_1\cdot e^{ax}+c_2\cdot e^{-ax}$
Where $\displaystyle c_1,c_2$ are constants (which can be chosen).
This is using the fact that any de with n solutions will also have solutions that are a linear combination of one or more of those n solutions, right? Would anyone know where there is a proof for this fact?

6. ## Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax) Originally Posted by Istafa How do i show this mathematically, preferably without using auxillary equations One of the most 'attracting' properties of the hyperboloc function is ...

$\displaystyle \frac{d}{dx} \sinh a x = a\ \cosh a x$ (1)

$\displaystyle \frac{d}{dx} \cosh a x = a\ \sinh a x$ (2)

... and from (1) and (2)...

$\displaystyle \frac{d^{2}}{d x^{2}} \sinh a x = a^{2}\ \sinh a x$ (3)

$\displaystyle \frac{d^{2}}{d x^{2}} \cosh a x = a^{2}\ \cosh a x$ (4)

... so that both $\displaystyle \sinh a x$ and $\displaystyle \cosh a x$ satisfy the ODE...

$\displaystyle \frac{d^{2} y}{d x^{2}} = a^{2}\ y$ (5)

But the (5) is a linear second order ODE and any linear combination of two linearly independent solutions is also a solution...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

7. ## Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

If the problem was, as initially stated, NOT to solve $\displaystyle y''- a^2y= 0$ but simply to show that y(x)= Asinh(ax)+ Bcosh(ax) is a solution, then you simply need to note that $\displaystyle (Acosh(ax)+ Bsinh(ax))''= Aa^2cosh(ax)+ Ba^2sinh(ax)$, as chisigma said, and that is $\displaystyle a^2(Acosh(ax)+ Bsinh(ax))= y$ so that $\displaystyle y''- a^2y= a^2y- a^2y= 0$.

By the way, just as sin(ax) and cos(ax) are solutions to the initial value problems $\displaystyle y''+ a^2y= 0$, y(0)= 0, y'(0)= 1 and $\displaystyle y''+ a^2y= 0$, y(0)= 1, y'(0)= 0, respectively, so sinh(ax) and cosh(ax) are solutions to the initial value problems $\displaystyle y''- a^2y= 0$, y(0)= 0, y'(0)= 1 and $\displaystyle y''- a^2y= 0$, y(0)= 1, y'(0)= 0, respectively.

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