superbeer Posted November 2, 2005 Share Posted November 2, 2005 Hi all, I'm trying to solve this differential equation but I'm getting strange results. Don't know how type all the symbols so I hope it makes sense. dy/dx + 3y = 6 dy/dx = 6 - 3y 1/(6 - 3y)(dy/dx) = 1 1/(6 -3y)dy = dx if I integrate both sides (-1/3 )ln[6-3y] = x + C boundary cond. when y=3 x=0 to find C, I substitute (-1/3)ln[6-3(3)] = C ok, this is where I'm stuck. Do I take C to be zero in this case. Also, does anyone know of a website where I can get a simple and easy introduction to first order linear differential equations? Thanks Link to comment Share on other sites More sharing options...
jcarlson Posted November 2, 2005 Share Posted November 2, 2005 (-1/3 )ln[6-3y] = x + C boundary cond. when y=3 x=0 to find C' date=' I substitute (-1/3)ln[6-3(3)'] = C ok, this is where I'm stuck. Do I take C to be zero in this case. ok, lets call this: (-1/3 )ln[6-3y] = x + C equation A and this: (-1/3)ln[6-3(3)] = C equation B. To get the value of C you solve the left side of equation B. Then you substitute the value of C into equation A, and you have your particular solution. Link to comment Share on other sites More sharing options...
superbeer Posted November 2, 2005 Author Share Posted November 2, 2005 ok' date=' lets call this: (-1/3 )ln[6-3y'] = x + C equation A and this: (-1/3)ln[6-3(3)] = C equation B. To get the value of C you solve the left side of equation B. Then you substitute the value of C into equation A, and you have your particular solution. Thanks for the reply The problem is the you can't take the natural logarithm of a negative number. Link to comment Share on other sites More sharing options...
cosine Posted November 3, 2005 Share Posted November 3, 2005 Thanks for the reply The problem is the you can't take the natural logarithm of a negative number. Would you like a complex C? In that case: ln(-3) = ln(-1) + ln(3) = i*pi + ln(3) Hope this helps. Link to comment Share on other sites More sharing options...
BobbyJoeCool Posted November 3, 2005 Share Posted November 3, 2005 ln(-1)=i? Link to comment Share on other sites More sharing options...
Dave Posted November 3, 2005 Share Posted November 3, 2005 That's not actually correct. You can see from the equation [imath]e^{\pi i} = -1[/imath] that ln(-1) can actually be any number [imath](2n+1)\pi i[/imath]. I think the more likely answer is that you've been given or wrongly written down an incorrect set of initial conditions. Link to comment Share on other sites More sharing options...
jcarlson Posted November 3, 2005 Share Posted November 3, 2005 Thanks for the reply The problem is the you can't take the natural logarithm of a negative number. And this is exactly why [math]\frac{d}{dx}\frac{1}{x} = ln(|x|)[/math] and not [math]ln(x)[/math] Link to comment Share on other sites More sharing options...
TD Posted November 3, 2005 Share Posted November 3, 2005 And this is exactly why [math]\frac{d}{dx}\frac{1}{x} = ln(|x|)[/math] and not [math']ln(x)[/math] Eh... [math]\frac{d}{dx}\frac{1}{x} = \frac{-1}{x^2}[/math]. [math]\int\frac{1}{x} dx = \ln|x|[/math] Link to comment Share on other sites More sharing options...
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