small problem

[superbeer]

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

[jcarlson]

(-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.

[superbeer]

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.

[cosine]

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.

[BobbyJoeCool]

ln(-1)=i?

[Dave]

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.

[jcarlson]

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]

[TD]

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]