# Explain $y''-y=0$

I've solved this equation and got $$c_0\cosh x + c_1\sinh x$$. However, I've noticed that if the arbitrary constant $$c_0$$ doesn't equal $$c_1$$, this wouldn't work; you would have to convert the $$\cosh x$$and $$\sinh x$$ into solutions of $$e^x$$ first. Why is this the case, because in other differential equations, the arbitrary constants can be any number?

• Use the substitution $$y(x)=e^{\lambda x}$$ – Dr. Sonnhard Graubner Dec 8 '18 at 14:55
• Why wouldn't it work? Your differential equation is satisfied for any values of $c_0$ and $c_1$, isn't it? – TonyK Dec 8 '18 at 15:00

Why do you claim that it doesn't work? $$(c_0\cosh +c_1\sinh )''=c_0(\sinh)'+c_1(\cosh)'=c_0\cosh+c_1\sinh$$

$$\cosh x$$ and $$\sinh x$$ are two independent linear combinations of $$e^x$$ and $$e^{-x}$$. Hence any linear combination of the two former can be expressed as a linear combination of the two latter, and conversely.

The hyperbolic functions are exponentials in disguise.

It works for any $$(c_0,c_1)$$. Since $$(\cosh(x))''=\cosh(x)$$ and $$(\sinh(x))''=\sinh(x),$$ if $$y=c_0 \cosh(x)+c_1\sinh(x),$$ then $$y''=c_0\cosh(x)+c_1\sinh(x)=y,$$ and so $$y''-y=0.$$

It is also true that $$\cosh(x)$$ and $$\sinh(x)$$ are L.I. functions, since the wronskian determinant is not zero, because $$W(\cosh(x),\sinh(x))=\left|\begin{matrix}\cosh(x)&\sinh(x)\\(\cosh(x))'&(\sinh(x))'\\\end{matrix}\right|=$$ $$=\left|\begin{matrix}\cosh(x)&\sinh(x)\\\sinh(x)&\cosh(x)\\\end{matrix}\right|=\cosh^2(x)-\sinh^2(x)=1,\quad \forall x.$$

This implies that all the solutions are of the form $$y=c_0 \cosh(x)+c_1\sinh(x).$$

• $\cosh'$ is $\sinh$, not $\cosh$. – Saucy O'Path Dec 8 '18 at 15:05
• Thanks, I forgot another prime on each expression. – Alejandro Nasif Salum Dec 8 '18 at 15:10