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

(-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.

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.

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.

ln(-1)=i?

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.

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]

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]