Solutions to differential equations

[nick.martinez]

y(x)=A*e^(λx) ; y'=λy

attempt at solution:

y'(x)= Ae^(λx)*λ

λy= Ae^(λx)*λ

divide by λ, which cancel. then i get:

y=Ae^(λx)

i want to say the differential equation holds but the issue i see is that y' and y'(x) are not equal derivatives, so my final answer is that the differential equation does not hold. what do you guys think?

 

[AugustCrawl]

$$ y=\text{Ae}^{\text{$\lambda $x}}\text{ } $$

 

[AugustCrawl]

Latex accident! Will repost later today :)

 

[HallsofIvy]

So you want to show that $y(x)= Ae^{\lambda x}$ is the general solution to the differential equation $y'= \lambda y$ or that it is a solution? The first is harder than the second! I don't understand what you mean when you say " y' and y'(x) are not equal derivatives". They are just slightly different ways of writing exactly the same thing- the first just doesn't have the "(x)" which is assumed since y is a function of x.

But your "proof" is a little over complicated. All you need is

If $y= Ae^{\lambda x}$ then $y'= A\lambda e^{\lambda x}= \lambda (A e^{\lambda x})= \lambda y$.

 

[nick.martinez]

So you want to show that $y(x)= Ae^{\lambda x}$ is the general solution to the differential equation $y'= \lambda y$ or that it is a solution? The first is harder than the second! I don't understand what you mean when you say " y' and y'(x) are not equal derivatives". They are just slightly different ways of writing exactly the same thing- the first just doesn't have the "(x)" which is assumed since y is a function of x.

But your "proof" is a little over complicated. All you need is

If $y= Ae^{\lambda x}$ then $y'= A\lambda e^{\lambda x}= \lambda (A e^{\lambda x})= \lambda y$.

when i say y' I am referring to the equation y'=λy and also refferring to y'(x)=Ae^(λx)*x in my example. i know they are both the same. Do the two equations i listed above look equal to each other?

 

[Mark44]

when i say y' I am referring to the equation y'=λy and also refferring to y'(x)=Ae^(λx)*x in my example.

There's a typo here. I think you mean y'(x) = Aλe^λx.

i know they are both the same. Do the two equations i listed above look equal to each other?

Equations are not "equal" to each other. Equations can be equivalent, which means they have the same set of solutions.

For your differential equation y' = λy, it can be shown that all solutions are of the form y = Ae^λx. If you differentiate the latter equation, you get y' = Aλe^λx = λy.

 

[nick.martinez]

There's a typo here. I think you mean y'(x) = Aλe^λx.
Equations are not "equal" to each other. Equations can be equivalent, which means they have the same set of solutions.

For your differential equation y' = λy, it can be shown that all solutions are of the form y = Ae^λx. If you differentiate the latter equation, you get y' = Aλe^λx = λy.

So in this case I can set y' and y'(x) equal to each other even though when I take the derivative of y(x) i get y'(x)=Ae^(λx)*λ which does not equal y' = λy? Again I did set them equal to each other and got the original y(x). So this is a solution to a diff eq.

 

[HallsofIvy]

You've got me really confused. If $y= Ae^{\lambda x}$ then $y'= A\lambda e^{\lambda x}= \lambda(Ae^{\lambda x})= \lambda y$ They are exactly the same!

 

[nick.martinez]

You've got me really confused. If $y= Ae^{\lambda x}$ then $y'= A\lambda e^{\lambda x}= \lambda(Ae^{\lambda x})= \lambda y$ They are exactly the same!

Thanks I wasn't thinking never mind. Thanks for the clarification.