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Solution of a Differential equation (Linear, 1st order). x=dependent variableby coolhand
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#1
Jun3012, 09:13 PM

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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
Jun3012, 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 [ 112xy^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? 


#3
Jun3012, 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)? 


#4
Jul112, 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? 


#5
Jul112, 12:23 AM

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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. 


#6
Jul112, 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.



#7
Jul112, 01:50 PM

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Rewrite [itex]\displaystyle \left [ 112xy^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] 


#8
Jul112, 05:38 PM

P: 15

So I just tried reworking through my algebra, starting with Sammy's suggest of rewriting the function as :
(dy/dx)=y^{3}+12xy^{2}(dy/dx) However, I had some confusion about multiplying through using y^{9}. 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 rewrote 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(e^{9})+y That doesn't seem right 


#9
Jul112, 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 


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Jul112, 07:15 PM

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#11
Jul112, 11:02 PM

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


#12
Jul312, 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 y^{12} as the integrating factor, and Q(x)=1/y^{3}: =[itex]\int[/itex]Q(y)P(y) =[itex]\int[/itex]y^{9}+C 5=(1/10)(2^{10})+C C=(487/5) x(y)=(1/P(y))Q(y)P(y)+C] x(y)=(y^{12}) [itex]\int[/itex][((1/10)y^{10})(487/5)] x(y)=(1/(10y^{2}))(487/(5y^{12})) 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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