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Calculus I: Solve this simple initial value problem.

G

Gerbil333

Diamond Member
============================
Solve the initial value problem:


dy/dx = x - 4 + xy - 4y

y(0) = 4
============================

My friend and I tried this and we got the same answer (see below), but it's incorrect. Where'd we go wrong?

dy/dx = 1(x-4) + y(x-4)

dy/dx = (y+1)(x-4)


~(1/(y+1))dy = ~(x-4)dx

("~" is supposed to be an integral symbol)

u = (y +1)
du = dy


~(1/u)dy = (x²/2) - 4x

ln(y+1) = (x²/2) - 4x

e^[ln(y+1)] = e^[(x²/2) - 4x]

y+1 = e^[(x²/2) - 4x]

y = e^[(x²/2) - 4x] - 1 + c

4 = e^[(0)²/2) - 4(0)] - 1 + c

4 = e^[0] - 1 + c

c = 4

y = e^[(x²/2) - 4x] + 3


So, it didn't seem that hard...what'd we do?
 
Originally posted by: Gerbil333
============================
Solve the initial value problem:


dy/dx = x - 4 + xy - 4y

y(0) = 4
============================

My friend and I tried this and we got the same answer (see below), but it's incorrect. Where'd we go wrong?

For easier typing purposes, I'm going to use ~ to represent the integral sign.


dy/dx = 1(x-4) + y(x-4)

dy/dx = (y+1)(x-4)

~(1/(y+1))dy = ~(x-4)dx

let u = (y +1)
du = dy

~(1/u)du = ((x^2)/2) - 4x

log(u) = x²/2 - 4x

u = 10^[x²/2 - 4x]

y+1 = 10^[x²/2 - 4x]

y= 10^[x²/2 - 4x] -1+c


4= 10^[0] -1+c

c=4

y= 10^[x²/2 - 4x] +3


So, it didn't seem that hard...what'd we do?
Click to expand...
log(x)!=ln(x) is it?

² is your friend
 
Thanks. I didn't know that worked in text boxes.

Edit:

Originally posted by: cirthix
log(x)!=ln(x)
FIXED
² is your friend
Click to expand...

How is that right? The derivative of ln(x) is (x'/x)

x' = 1
x = x

So, (1/x) is the derivative of ln(x)

No?
 
hehe, yeah, but for some reason it doesnt wanna work over aim sometimes 😕
is it the right answer?
 
dy/dx = (x-4)*(y+1)

dy/(y+1) = dx*(x-4)

ln(y+1) = x^2-4x+C
y+1 = A*e^(x^2/2-4x)
This is where you went wrong. When you exponentiate the constant C, you have a constant scalar out in front. I called it A.

Since y(0)=4, then

A=5, so

y(x) = 5*e^(x^2/2-4x)-1
 
Originally posted by: chuckywang
dy/dx = (x-4)*(y+1)

dy/(y+1) = dx*(x-4)

ln(y+1) = x^2-4x+C
y+1 = A*e^(x^2/2-4x)
This is where you went wrong. When you exponentiate the constant C, you have a constant scalar out in front. I called it A.

Since y(0)=4, then

A=5, so

y(x) = 5*e^(x^2/2-4x)-1
Click to expand...

:thumbsup:
 
Originally posted by: chuckywang
dy/dx = (x-4)*(y+1)

dy/(y+1) = dx*(x-4)

ln(y+1) = x^2-4x+C
y+1 = A*e^(x^2/2-4x)
This is where you went wrong. When you exponentiate the constant C, you have a constant scalar out in front. I called it A.

Since y(0)=4, then

A=5, so

y(x) = 5*e^(x^2/2-4x)-1
Click to expand...

Ah, ok. My book failed to explain that (high school calc book). I hope the college books I'll have next year won't suck like this one does. I have an A right now, but only because I've used alternate learning sources.
 
Originally posted by: Gerbil333
Originally posted by: chuckywang
dy/dx = (x-4)*(y+1)

dy/(y+1) = dx*(x-4)

ln(y+1) = x^2-4x+C
y+1 = A*e^(x^2/2-4x)
This is where you went wrong. When you exponentiate the constant C, you have a constant scalar out in front. I called it A.

Since y(0)=4, then

A=5, so

y(x) = 5*e^(x^2/2-4x)-1
Click to expand...

Ah, ok. My book failed to explain that (high school calc book). I hope the college books I'll have next year won't suck like this one does. I have an A right now, but only because I've used alternate learning sources.
Click to expand...

like ATOT?
 
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