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

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In summary, the given function is a solution of the differential equation y''-y=0, y_1(x)=e^x, y_2(x)=\cosh{x}, and $y_3=c_1e^x+c_2e^{-x}$.
  • #1
karush
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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
 
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  • #2
karush said:
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:

\(\displaystyle y_n''-y_n=0\) where \(\displaystyle n\in\{1,2\}\)

In other words, are the given functions equal to their own second derivative?
 
  • #3
MarkFL said:
You want to see if:

\(\displaystyle y_n''-y_n=0\) where \(\displaystyle n\in\{1,2\}\)

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
 
  • #4
karush said:

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

Yes:

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

What about the hyperbolic cosine function $y_2(x)$ ?
 
  • #5
Don't know never did em.

Guess I cud W|A
 
  • #6
karush said:
Don't know never did em.

Guess I cud W|A

Recall:

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

How about:

\(\displaystyle y_3(x)=c_1e^x+c_2e^{-x}\) ?
 
  • #7
MarkFL said:
Recall:

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

How about:

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

No memories of it
I'll have to look it up
 
  • #8
karush said:
No memories of it
I'll have to look it up

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

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

What about $y_3$?
 
  • #9
$ 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 $\frac{d^2}{dx^2}$

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

$\color{blue}{\tiny\textbf{Elementary Differential Equations and Boundary Value Problems}}$
$\color{blue}{\tiny\textsf{William E. Boyce, Edward P. Hamilton and Richard C. DiPrima }}$
 
Last edited:
  • #10
MarkFL said:
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)
 
  • #11
it wasn't asked!
 
  • #12
karush said:
it wasn't asked!

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

MarkFL said:
How about:

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

MarkFL said:
What about $y_3$?

MarkFL said:
MarkFL said:
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)
 
  • #13
$\displaystyle y''-y=0\\
y_1(x)=e^x\\
y_2(x)=\cosh{x}\\
y_3(x)=c_1e^x+c_2e^{-x}\\
( c_1e^x+c_2e^{-x} )''-c_1e^x+c_2e^{-x} =\\
( c_1e^x+c_2e^{-x} )-( c_1e^x+c_2e^{-x} )=0$
 

FAQ: Verify Solutions of Differential Equation y''-y=0

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It can be used to model many real-world phenomena in physics, engineering, and other fields.

What does it mean to verify a solution of a differential equation?

Verifying a solution of a differential equation means to substitute the proposed solution into the equation and show that it satisfies the equation for all values of the independent variable.

Why is it important to verify solutions of differential equations?

It is important to verify solutions of differential equations because it ensures that the proposed solution is a valid and accurate representation of the system being modeled. It also allows for the identification of any errors or mistakes in the solution.

How do you verify a solution of a differential equation?

To verify a solution of a differential equation, substitute the proposed solution into the equation and then simplify using mathematical operations. If the resulting equation is true for all values of the independent variable, then the solution is verified.

Can a differential equation have multiple solutions?

Yes, a differential equation can have multiple solutions. In fact, most differential equations have an infinite number of solutions. This is because there are often many different functions that can satisfy the equation and accurately model the system being studied.

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