MHB Verify Solutions of Differential Equation y''-y=0

[karush]

Verify for each of the following that the given function is a
solution of the differentual equation
$$\displaystyle y''-y=0 \quad y_1(x)=e^x \quad y_2(x)=\cosh{x}$$

ok I don't understand the question exactly
the first only looks like possibility

 

[MarkFL]

Verify for each of the following that the given function is a
solution of the differentual equation
$$\displaystyle y''-y=0 \quad y_1(x)=e^x \quad y_2(x)=\cosh{x}$$

ok I don't understand the question exactly
the first only looks like possibility

You want to see if:

$$y_n''-y_n=0$$ where $$n\in\{1,2\}$$

In other words, are the given functions equal to their own second derivative?

 

[karush]

​

You want to see if:

$$y_n''-y_n=0$$ where $$n\in\{1,2\}$$

In other words, are the given functions equal to their own second derivative?

would that be $y_1$ since $e^x$ is the function and its derivatives are always the same

 

[MarkFL]

​

would that be $y_1$ since $e^x$ is the function and its derivatives are always the same

Yes:

$$\frac{d^2}{dx^2}\left(e^x\right)=e^x$$

What about the hyperbolic cosine function $y_2(x)$ ?

 

[karush]

Don't know never did em.

Guess I cud W|A

 

[MarkFL]

Don't know never did em.

Guess I cud W|A

Recall:

$$\cosh(x)\equiv\frac{e^x+e^{-x}}{2}$$

How about:

$$y_3(x)=c_1e^x+c_2e^{-x}$$ ?

 

[karush]

Recall:

$$\cosh(x)\equiv\frac{e^x+e^{-x}}{2}$$

How about:

$$y_3(x)=c_1e^x+c_2e^{-x}$$ ?

No memories of it
I'll have to look it up

 

[MarkFL]

No memories of it
I'll have to look it up

You can do the differentiation on the exponential definition of the function...

$$\cosh(x)=\frac{1}{2}\left(e^x+e^{-x}\right)$$

What about $y_3$?

 

[karush]

$ y''-y=0;\\
\displaystyle\exp\int \, dx = e^x
\implies e^x(y''-y)=0
\implies e^x-e^x=0\\
y_1(x)=e^x
\implies (e^x)''-(e^x)=0
\implies (e^x)-(e^x)=0\\
y_2(x)=\cosh{x}
\implies (\cosh{x})''-(\cosh{x})=0
\implies (\cosh{x})-(\cosh{x})=0$\\

assuming $\frac{d^2}{dx^2}$

$\displaystyle \cosh(x)''
=\left(\frac{1}{2}\left(e^x+e^{-x}\right)\right)''
=\frac{1}{2}\left(e^x+e^{-x}\right)$ \\ \\

$\color{blue}{\tiny\textbf{Elementary Differential Equations and Boundary Value Problems}}$
$\color{blue}{\tiny\textsf{William E. Boyce, Edward P. Hamilton and Richard C. DiPrima }}$

 

[MarkFL]

What about $y_3$?

I guess this question has gotten lost in all the excitement the first two times I asked it? Hopefully it is visible now. (Giggle)

 

[karush]

it wasn't asked!

 

[MarkFL]

it wasn't asked!

I asked it a total of 3 times. You kept ignoring me for some reason.

How about:

$$y_3(x)=c_1e^x+c_2e^{-x}$$ ?

What about $y_3$?

What about $y_3$?

I guess this question has gotten lost in all the excitement the first two times I asked it? Hopefully it is visible now. (Giggle)

 

[karush]

$\displaystyle y''-y=0\\
y_1(x)=e^x\\
y_2(x)=\cosh{x}\\
y_3(x)=c_1e^x+c_2e^{-x}\\
( c_1e^x+c_2e^{-x} )''-c_1e^x+c_2e^{-x} =\\
( c_1e^x+c_2e^{-x} )-( c_1e^x+c_2e^{-x} )=0$