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Reduction of Order no independent or dependent variable

  • Thread starter gremmie
  • Start date
  • #1
4
0

Homework Statement



y'' +(y')^2 = 6(y') -9

Homework Equations


u = y'
u' = y''


The Attempt at a Solution



equation from 1 becomes

u' + U^2 = 6u - 9

du/dx + u ^2 = 6u-9

du/dx = -u^2 + 6u -9

du/dx = -(u^2 +6u - 9)

du/dx = -(u-3)^2

du/(u-3)^2 = - dx

integrate both sides

1/(u-3) = x

sub in dy/dx

1/(dy/dx -3) = x

from this point ive tried multiple ways to get dy and dx seperated but im not doing something right... thanks for the help

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
33,646
5,313

Homework Statement



y'' +(y')^2 = 6(y') -9

Homework Equations


u = y'
u' = y''


The Attempt at a Solution



equation from 1 becomes

u' + U^2 = 6u - 9

du/dx + u ^2 = 6u-9

du/dx = -u^2 + 6u -9

du/dx = -(u^2 +6u - 9)
Typo above. It should be -(u^2 - 6u + 9)
du/dx = -(u-3)^2

du/(u-3)^2 = - dx

integrate both sides

1/(u-3) = x
You're missing the constant of integration. Also, instead of substituting dy/dx as you're doing in the next step, solve for u, then substitute.
sub in dy/dx

1/(dy/dx -3) = x

from this point ive tried multiple ways to get dy and dx seperated but im not doing something right... thanks for the help

Homework Statement





Homework Equations





The Attempt at a Solution

 
  • #3
4
0
Ah yep sure did have a typo ...thanks ....sorry first time posting....yep i left my constant of integration off as well..

Ok so picking up where i left off but this time solving for u
1/(u-3) = x +C

1 = (u-3)(x+C)
ux+uc = 1 + 3C +3x
u(x+C) = 1+3C +3x

u = (1+3C +3x)/(x+C)
sub for u
dy/dx = (1+3C +3x)/(x+C)

dy = (1+3C +3x)/(x+C) dx

u sub u = x+c
du = dx

Integral (3u+1)/u

s sub
s = 3u+1
ds = 3du

integral s/(s-1)

integral 1 + 1/(s-1)

integral 1 +integral 1/(s-1)

p sub for 1/(s-1)

thus integral dp/p

evaluate integrals

s +lnp + C2(2nd constant of integration)

sub back

s + ln(s-1) +C2

3u + 1 +ln(3u+1-1) +C2
3u + 1 +ln(3u) +C2

3(x+C) +1 +ln(3x +3C) +C2
so
y= 3x + 3C +1 +ln(3x+ 3C) +C2

the answer listed is : y = 3x +ln(x+C) +C2
So i see that im on the right track ... but im still missing something again .. thanks
 
  • #4
33,646
5,313
Ah yep sure did have a typo ...thanks ....sorry first time posting....yep i left my constant of integration off as well..

Ok so picking up where i left off but this time solving for u
1/(u-3) = x +C

1 = (u-3)(x+C)
ux+uc = 1 + 3C +3x
u(x+C) = 1+3C +3x
You sure took the long way around. Starting from your first equation,
1/(u-3) = x +C
==> u - 3 = 1/(x + C)
==> u = 3 + 1/(x + C)

Undoing the substitution,
dy/dx = 3 + 1/(x + C)
==> dy = [3 + 1/(x + C)]dx
==> y = 3x + ln|x + C| + D

u = (1+3C +3x)/(x+C)
sub for u
dy/dx = (1+3C +3x)/(x+C)

dy = (1+3C +3x)/(x+C) dx

u sub u = x+c
du = dx

Integral (3u+1)/u

s sub
s = 3u+1
ds = 3du

integral s/(s-1)

integral 1 + 1/(s-1)

integral 1 +integral 1/(s-1)

p sub for 1/(s-1)

thus integral dp/p

evaluate integrals

s +lnp + C2(2nd constant of integration)

sub back

s + ln(s-1) +C2

3u + 1 +ln(3u+1-1) +C2
3u + 1 +ln(3u) +C2

3(x+C) +1 +ln(3x +3C) +C2
so
y= 3x + 3C +1 +ln(3x+ 3C) +C2

the answer listed is : y = 3x +ln(x+C) +C2
So i see that im on the right track ... but im still missing something again .. thanks
[/quote]
 
  • #5
4
0
ah man i sure did... Well thanks for your help! I knew there had to be an easier way! everyone in class got this problem done in class an ive been messing with it for about 2 weeks! Again, Thank You.
 

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