BAH! Calculus Question, again, and again. Ridiculous.

[Turkish]

I got a problem here. This is Calculus 3: Applications of Definite Stability.

The problem is pretty easy actually. I have done tons of these. But for some reason I can't get the answer for this one.

Here is the data given.

y'=3y-2t and y(2)= -2
Estimate y(3) using Euler's Numerical Method (h=1)

Euler's Numerical Method is y(x+1)= y(x)+(h)(y'(x))

Here is how I do it:

y(3)= y(2)+(1)(y'(2))
y(3)= -2+(1)(3(2)-2t)
y(3)= -2+6-2t
y(3)= 4-2t

To find t, I just took calculated the integral for y'=3y-2t, so that is (3y^2)/2 - 2ty
Since I know y(2)= -2, I just plugged in 2 for y, and got (3*4)/2 - 4t = -2
That gave me t= -2

I found y(3)= 4-2t earlier, so I plugged in 2 for t and got y(3)=0.

Well, in the solutions manual, the answer is frickin "-12". I don't know what I am doing wrong.

Can you Math gurus help me out here? I really feel sad for not being able to solve such an easy question.

Thanks,
Xiety

 

[MaxFusion16]

i mean you should've plugged in -2 for y and 2 for t
the correct formula is y1=y0+h*F(x0,y0)

 

[Gibson486]

i think you took the integral wrong. It's a function of t, not y.

edit. not teh itegral, i mean you plugged in your givens wrong. I nother words, for the solution, you plug in for t, not y.

""To find t, I just took calculated the integral for y'=3y-2t, so that is (3y^2)/2 - 2ty
Since I know y(2)= -2, I just plugged in 2 for y, and got (3*4)/2 - 4t = -2
That gave me t= -2""

Sorry, I am not clear, buts it 1:30 and diffy q's is making me cry

 

[Turkish]

Originally posted by: Gibson486
i think you took the integral wrong. It's a function of t, not y.

edit. not teh itegral, i mean you plugged in your givens wrong. I nother words, for the solution, you plug in for t, not y.

""To find t, I just took calculated the integral for y'=3y-2t, so that is (3y^2)/2 - 2ty
Since I know y(2)= -2, I just plugged in 2 for y, and got (3*4)/2 - 4t = -2
That gave me t= -2""

Sorry, I am not clear, buts it 1:30 and diffy q's is making me cry

Hmmm. Well even if I plugged them the other way around, it still doesn't give me "-12".

Any other suggestions?

 

[Gibson486]

Also, when you take an integral, you cant forget about the constant. The Y(2)=-2 is to solve for the constant, not t.

 

[Turkish]

Originally posted by: Gibson486
Also, when you take an integral, you cant forget about the constant. The Y(2)=-2 is to solve for the constant, not t.

maybe I am not clear either but what you tell me doesn't make any difference to the final outcome. it changes the result but doesn't give me the correct answer.

 

[Gibson486]

NM, but eulers comes back to haunt you (it haunts me) in differential equations, so be ready for it.

 

[MaxFusion16]

no need to integrate, here is how it's done
y(3) = -2 + 1*(3(-2) - 2(2))
y(3) = -12

 

[Gibson486]

""maybe I am not clear either but what you tell me doesn't make any difference to the final outcome. it changes the result but doesn't give me the correct answer. ""

They are all steps in the right direction (You NEED the constant when you integrate most of the time and you cant plug in for Y if it is a function of T), but read my above post. I just noticed this also, it's eulers, so just use teh function as is.

 

[Turkish]

Originally posted by: MaxFusion16
no need to integrate, here is how it's done
y(3) = -2 + 1*(3(-2) - 2(2))
y(3) = -12

Why are you putting a -2 there? We are pluggin in the value for y(2) = -2 where the value is 2. Not -2.

 

[Turkish]

Originally posted by: Gibson486
""maybe I am not clear either but what you tell me doesn't make any difference to the final outcome. it changes the result but doesn't give me the correct answer. ""

They are all steps in the right direction (You NEED the constant when you integrate most of the time and you cant plug in for Y if it is a function of T), but read my above post. I just noticed this also, it's eulers, so just use teh function as is.

Well you need to integrate the derivative in order to find the constant value, whether it is y or t.

 

[MaxFusion16]

Originally posted by: Xiety

Originally posted by: MaxFusion16
no need to integrate, here is how it's done
y(3) = -2 + 1*(3(-2) - 2(2))
y(3) = -12

Why are you putting a -2 there? We are pluggin in the value for y(2) = -2 where the value is 2. Not -2.

because the function is 3y-2t and y(2) = -2
so plug in y for y and t for t

 

[Gibson486]

Why are you putting a -2 there? We are pluggin in the value for y(2) = -2 where the value is 2. Not -2.

y at 2 is -2.

 

[Turkish]

Originally posted by: MaxFusion16

Originally posted by: Xiety

Originally posted by: MaxFusion16
no need to integrate, here is how it's done
y(3) = -2 + 1*(3(-2) - 2(2))
y(3) = -12

Why are you putting a -2 there? We are pluggin in the value for y(2) = -2 where the value is 2. Not -2.

because the function is 3y-2t and y(2) = -2
so plug in y for y and t for t

LoL, but y is not (-2), it's (2)!

 

[Gibson486]

no, y is -2

Y(2)=-2. remember Y(t)=t. That is why i said it was a function of t, not y. Whatever is on that parethesis is the variable you plug into. The -2 tell syou that after you plug in for teh function, you should get that value for y.

 

[Turkish]

Originally posted by: Gibson486
no, y is -2

Y(2)=-2. remember Y(t)=t. That is why i said it was a function of t, not y. Whatever is on that parethesis is the variable you plug into. The -2 tell syou that after you plug in for teh function, you should get that value for y.

ohhhh i think i got your point now!

Hmmmm well, thanks!

I dunno, I think i was braindead.

:beer: for Gibson and MaxFusion.

 

[Gibson486]

I think what is throwing you off is how yoru wrote yoru equation. some of those numbers are subscripts, not conditions.

Euler formula

x (subscript x-1) = x (subscript x) * h(f(x,y,z...etc))