split - Please Help With Euler's Method?

[VictimOfCalculus]

Thank you!

Could I throw another one out there? It's similar, but I can't get it to work so smoothly;

Consider the initial value problem y[sup:3qn5fgrb]I[/sup:3qn5fgrb]=1-y/x, y(1)=1
Verify that y=(x[sup:3qn5fgrb]2[/sup:3qn5fgrb]+1)/(2x) is the solution to this IVP

So I have;

y=(x[sup:3qn5fgrb]2[/sup:3qn5fgrb]+1)/(2x)
y[sup:3qn5fgrb]I[/sup:3qn5fgrb]=(2x)(2x)-(x2+1)(2)/(2x)[sup:3qn5fgrb]2[/sup:3qn5fgrb]
y[sup:3qn5fgrb]I[/sup:3qn5fgrb]=4x[sup:3qn5fgrb]2[/sup:3qn5fgrb]-2x[sup:3qn5fgrb]2[/sup:3qn5fgrb]+2/4x[sup:3qn5fgrb]2[/sup:3qn5fgrb]=(x[sup:3qn5fgrb]2[/sup:3qn5fgrb]+1)/2x[sup:3qn5fgrb]2[/sup:3qn5fgrb]=y/x

So I'm missing the (1-) part

The second part I think I've got;

y(1)=(1[sup:3qn5fgrb]2[/sup:3qn5fgrb]+1)/(2(1))=1 (Satisfies initial condition)

Thank you! =]

-Amy

 

[Deleted member 4993]

Re: Please Help With Euler's Method?

Thank you!

Could I throw another one out there? It's similar, but I can't get it to work so smoothly;

Consider the initial value problem y[sup:1mbklunm]I[/sup:1mbklunm]=1-y/x,

Is that y' =(1-y)/x or y' = 1 - (y/x)......... need to watch those grouping symbols

y(1)=1
Verify that y=(x[sup:1mbklunm]2[/sup:1mbklunm]+1)/(2x) is the solution to this IVP

So I have;

y=(x[sup:1mbklunm]2[/sup:1mbklunm]+1)/(2x)
y[sup:1mbklunm]I[/sup:1mbklunm]=[(2x)(2x)-(x[sup:1mbklunm]2[/sup:1mbklunm]+1)(2)]/(2x)[sup:1mbklunm]2[/sup:1mbklunm] .......watch those grouping symbols
y[sup:1mbklunm]I[/sup:1mbklunm] = [4x[sup:1mbklunm]2[/sup:1mbklunm]-2x[sup:1mbklunm]2[/sup:1mbklunm]-2]/[4x[sup:1mbklunm]2[/sup:1mbklunm]] = (x[sup:1mbklunm]2[/sup:1mbklunm]-1)/[2x[sup:1mbklunm]2[/sup:1mbklunm]] = 1 - (y/x)

y/x

So I'm missing the (1-) part

The second part I think I've got;

y(1)=(1[sup:1mbklunm]2[/sup:1mbklunm]+1)/(2(1))=1 (Satisfies initial condition)

Thank you! =]

-Amy

Please start a new thread for a new problem