Solution of a Differential equation (Linear, 1st order). x=dependent variable


by coolhand
Tags: differential eqns
coolhand
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#1
Jun30-12, 09:13 PM
P: 15
the problem is as follows:

[1-(12*x*(y^2))]*(dy/dx)=y^3

we need to solve this equation such as x=5 & y=2 by regarding y as the independent variable.

Here is my work for the first attempt:

Step 1)separation & integration
Step 2)ln(x)= (-1/(2y^2))-12ln(y)+C
x=Ce^(-1/(2y^2)) +(1/(y^12))
x(y)=5.665

anyone know where I went wrong?
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Simon Bridge
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#2
Jun30-12, 09:36 PM
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I'll translate your text to atex and you tell me if I've understood you:
You started with the following to solve...
[tex]\left [ 1-12xy^2 \right ] \frac{dy}{dx} = y^3[/tex]...but you want the solution in the form [itex]x=f(y)[/itex]...you can solve for y=f(x) and change the variable or figure how dy/dx is related to dx/dy.

Anyway - you did it the first way and you are pretty sure of your solution:
[tex]\ln{(x)} = -\frac{1}{2y^2} - 12\ln{(y)} + c[/tex]

Taking the exponential of both sides gives me:
[itex]x = C \exp{ \big [-\frac{1}{2y^2} - 12\ln{(y)}\big ]}[/itex] which is:
[itex]x = C \exp{ [-\frac{1}{2y^2}]}\exp{[\ln{(y^{-12})}]}[/itex]... does this make sense?
coolhand
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#3
Jun30-12, 10:20 PM
P: 15
Simon,

that makes sense, we essentially came to the same solution of the general equation. However, when I plug in for the given x & y values to solve for "C", I get a huge number on the order of 2*10^4


to interconvert between dy/dx, can you just simply invert (given that you invert the functions as well)?

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Jul1-12, 12:03 AM
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Solution of a Differential equation (Linear, 1st order). x=dependent variable


What I got was different to your result of x=Ce^(-1/(2y^2)) +(1/(y^12)) which I read as:

[tex]x = C\exp{\left [ -\frac{1}{2y^2} + \frac{1}{y^{12}} \right ]}[/tex] which tells me that [tex]C=x\exp{\left [ \frac{1}{2y^2} - \frac{1}{y^{12}} \right ]}[/tex] ... when I put in x=5 and y=2, I get C=36.936.

Mine is different - it's the y^12 that throws it out ... and I get about C=30,000.
... always assuming the solution you got is correct. Any reason to suspect C is not big?
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Jul1-12, 12:23 AM
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Quote Quote by coolhand View Post
the problem is as follows:

[1-(12*x*(y^2))]*(dy/dx)=y^3

we need to solve this equation such as x=5 & y=2 by regarding y as the independent variable.

Here is my work for the first attempt:

Step 1)separation & integration
Step 2)ln(x)= (-1/(2y^2))-12ln(y)+C
x=Ce^(-1/(2y^2)) +(1/(y^12))
x(y)=5.665

anyone know where I went wrong?
It looks like your solution to the differential equation is wrong.

Taking the derivative of both side of [itex]\displaystyle \ln{(x)} = -\frac{1}{2y^2} - 12\ln{(y)} + c[/itex] gives:

[itex]\displaystyle \frac{1}{x} = 2\frac{1}{2y^3}y' - 12\frac{1}{y}y' [/itex]

and finally [itex]\displaystyle y^3 = \left(x - 12xy^3\right)\frac{dy}{dx} [/itex]

It doesn't look like your D.E. is separable.

Find an integrating factor to make it exact.
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Jul1-12, 01:50 AM
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I think the y^12 indicates that an integrating factor has been tried... but when I tried it I got a different solution. Advise to OP: revisit the differential equation. Check the algebra.
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Jul1-12, 01:50 PM
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Rewrite [itex]\displaystyle \left [ 1-12xy^2 \right ] \frac{dy}{dx} = y^3[/itex]

as [itex]\displaystyle \frac{dy}{dx}=y^3+12xy^2\,\frac{dy}{dx}\,.[/itex]

Multiply through by [itex]y^9\,.[/itex]

[itex]\displaystyle \frac{d}{dx}\left(x\cdot f(x)\right)=f(x)+x\cdot\frac{d}{dx}\left(f(x) \right)[/itex]
coolhand
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#8
Jul1-12, 05:38 PM
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So I just tried reworking through my algebra, starting with Sammy's suggest of rewriting the function as :

(dy/dx)=y3+12xy2(dy/dx)

However, I had some confusion about multiplying through using y9. Is that step generating an integrating factor such that:

y= [([itex]\int[/itex]Q(x)*M(x)dx)]/(M(x))
then, P(x)= [M(x)']/M(x)
then the integrating factor would be e^([itex]\int[/itex]P(x))?

If so, I was confused as to which function would be the appropriate Q(x), in the form that you re-wrote it in, it seemed to be y^3. However, I could be completely mistaken.

Using y^3 as Q(x), I calculated P(x)=e^([itex]\int[/itex](9/y))
The integrating factor was then:

C(e9)+y

That doesn't seem right
Simon Bridge
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Jul1-12, 06:55 PM
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Aside: Get into the habit: what is it about the function that leads you to think "that doesn't seem right"? You get a gut reaction: "ugh - somefink rong - tell brain", try investigating to figure what triggered the reaction.

[itex]y^9[/itex] is the integrating factor. Multiply through to give:


[tex]y^9 \frac{dy}{dx} = y^{12} + 12xy^{11} \frac{dy}{dx}[/tex]... then identify P(x,y) and Q(x,y).

Hint: Try rearranging so that it has form: P(x,y)+Q(x,y)y' = 0
HallsofIvy
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Jul1-12, 07:15 PM
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Quote Quote by coolhand View Post
the problem is as follows:

[1-(12*x*(y^2))]*(dy/dx)=y^3

we need to solve this equation such as x=5 & y=2 by regarding y as the independent variable.

Here is my work for the first attempt:

Step 1)separation & integration
You are missing the most important point if you want us to point out a mistake: how did you separate? This looks to me like it is not a separable equation.

Step 2)ln(x)= (-1/(2y^2))-12ln(y)+C
x=Ce^(-1/(2y^2)) +(1/(y^12))
x(y)=5.665

anyone know where I went wrong?
As you said in your title, this is a linear equation. Writing it as a function x, in terms of variable y, it is [itex]dx/dy+ 12y^2x= 1/y[/itex]. There is a standard formula for the integrating factor of a linear equation.
SammyS
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Jul1-12, 11:02 PM
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Quote Quote by HallsofIvy View Post
...

As you said in your title, this is a linear equation. Writing it as a function x, in terms of variable y, it is [itex]dx/dy+ 12y^2x= 1/y[/itex]. There is a standard formula for the integrating factor of a linear equation.
I get [itex]\displaystyle \left(y^3\right)\frac{dx}{dy}+ 12y^2x= 1\ .[/itex]

This is equivalent to [itex]\displaystyle \frac{dx}{dy}+ 12\frac{x}{y}= \frac{1}{y^3}\ .[/itex]
coolhand
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#12
Jul3-12, 12:08 AM
P: 15
Ok, so after reworking this problem a few more times, I think I'm finally close the right solution:

using y12 as the integrating factor, and Q(x)=1/y3:

=[itex]\int[/itex]Q(y)P(y)
=[itex]\int[/itex]y9+C

5=(1/10)(210)+C
C=(-487/5)

x(y)=(1/P(y))Q(y)P(y)+C]
x(y)=(y-12) [itex]\int[/itex][((1/10)y10)-(487/5)]
x(y)=(1/(10y2))-(487/(5y12))


however when I input this as an answer it was wrong. Should I have solved for C later in the problem (i.e. right before my final answer)?


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