Reduction of Order no independent or dependent variable

In summary, I am still missing something and I am not sure what it is. Can you help?[/quote]In summary, you are still missing something in your solution to the homework equation. You are welcome for help.
  • #1
gremmie
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 I've tried multiple ways to get dy and dx separated but I am not doing something right... thanks for the help
 
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  • #2
gremmie said:

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)
gremmie said:
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.
gremmie said:
sub in dy/dx

1/(dy/dx -3) = x

from this point I've tried multiple ways to get dy and dx separated but I am not doing something right... thanks for the help
 
  • #3
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 I am on the right track ... but I am still missing something again .. thanks
 
  • #4
gremmie said:
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

gremmie said:
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 I am on the right track ... but I am still missing something again .. thanks

[/quote]
 
  • #5
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 I've been messing with it for about 2 weeks! Again, Thank You.
 

Related to Reduction of Order no independent or dependent variable

1. What is reduction of order in scientific terms?

Reduction of order is a mathematical technique used to find a second solution to a differential equation by reducing its order. This method is commonly used in physics and engineering to solve complex equations with multiple variables.

2. How does reduction of order work?

Reduction of order involves substituting a new variable into the original equation, which reduces the order of the equation by one. This new variable is typically a function of the original independent variable, and by solving for it, we can find a second solution to the differential equation.

3. What is the significance of reduction of order in scientific research?

Reduction of order is an important tool in scientific research as it allows for the simplification of complex differential equations. By finding a second solution to an equation, we can gain a better understanding of the behavior and relationships between variables in a system.

4. Can reduction of order be used for any type of differential equation?

No, reduction of order can only be used for certain types of differential equations, specifically those that are linear and homogeneous. This means that the dependent variable and its derivatives appear only in linear combinations in the equation.

5. Are there any limitations or drawbacks to using reduction of order?

One limitation of reduction of order is that it only provides a second solution to a differential equation, which may not be enough to fully understand the behavior of a system. Additionally, this method can only be used for equations with specific characteristics, making it limited in its applicability.

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