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

1. Jun 30, 2012

### coolhand

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?

2. Jun 30, 2012

### Simon Bridge

I'll translate your text to atex and you tell me if I've understood you:
You started with the following to solve...
$$\left [ 1-12xy^2 \right ] \frac{dy}{dx} = y^3$$...but you want the solution in the form $x=f(y)$...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:
$$\ln{(x)} = -\frac{1}{2y^2} - 12\ln{(y)} + c$$

Taking the exponential of both sides gives me:
$x = C \exp{ \big [-\frac{1}{2y^2} - 12\ln{(y)}\big ]}$ which is:
$x = C \exp{ [-\frac{1}{2y^2}]}\exp{[\ln{(y^{-12})}]}$... does this make sense?

3. Jun 30, 2012

### coolhand

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)?

4. Jul 1, 2012

### Simon Bridge

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

$$x = C\exp{\left [ -\frac{1}{2y^2} + \frac{1}{y^{12}} \right ]}$$ which tells me that $$C=x\exp{\left [ \frac{1}{2y^2} - \frac{1}{y^{12}} \right ]}$$ ... 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?

Last edited: Jul 1, 2012
5. Jul 1, 2012

### SammyS

Staff Emeritus
It looks like your solution to the differential equation is wrong.

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

$\displaystyle \frac{1}{x} = 2\frac{1}{2y^3}y' - 12\frac{1}{y}y'$

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

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

Find an integrating factor to make it exact.

Last edited: Jul 1, 2012
6. Jul 1, 2012

### Simon Bridge

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.

7. Jul 1, 2012

### SammyS

Staff Emeritus
Rewrite $\displaystyle \left [ 1-12xy^2 \right ] \frac{dy}{dx} = y^3$

as $\displaystyle \frac{dy}{dx}=y^3+12xy^2\,\frac{dy}{dx}\,.$

Multiply through by $y^9\,.$

$\displaystyle \frac{d}{dx}\left(x\cdot f(x)\right)=f(x)+x\cdot\frac{d}{dx}\left(f(x) \right)$

8. Jul 1, 2012

### coolhand

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= [($\int$Q(x)*M(x)dx)]/(M(x))
then, P(x)= [M(x)']/M(x)
then the integrating factor would be e^($\int$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^($\int$(9/y))
The integrating factor was then:

C(e9)+y

That doesn't seem right

9. Jul 1, 2012

### Simon Bridge

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.

$y^9$ is the integrating factor. Multiply through to give:

$$y^9 \frac{dy}{dx} = y^{12} + 12xy^{11} \frac{dy}{dx}$$... 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

10. Jul 1, 2012

### HallsofIvy

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.

As you said in your title, this is a linear equation. Writing it as a function x, in terms of variable y, it is $dx/dy+ 12y^2x= 1/y$. There is a standard formula for the integrating factor of a linear equation.

11. Jul 1, 2012

### SammyS

Staff Emeritus
I get $\displaystyle \left(y^3\right)\frac{dx}{dy}+ 12y^2x= 1\ .$

This is equivalent to $\displaystyle \frac{dx}{dy}+ 12\frac{x}{y}= \frac{1}{y^3}\ .$

Last edited: Jul 1, 2012
12. Jul 3, 2012

### coolhand

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:

=$\int$Q(y)P(y)
=$\int$y9+C

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

x(y)=(1/P(y))Q(y)P(y)+C]
x(y)=(y-12) $\int$[((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)?

Last edited: Jul 3, 2012