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How do I solve this differential eqn?

1. The problem statement, all variables and given/known data
integrate:

##-v(du/dy) = κ(d^2(u)/dy^2) ##

to obtain:

## (-v/κ)y = ln(du/dy) + c##

and finally:

##u = d + w*e^(-vy/κ)##

2. Relevant equations
##-v(du/dy) = κ(d^2(u)/dy^2) ##
## (-v/κ)y = ln(du/dy) + c##

3. The attempt at a solution
## (-v/κ)dy = d(u) ##

which gives:

## (-vy/κ) + C = u ##
 

Ssnow

Gold Member
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144
##v,\kappa## are constant ? What is ##w## ?

I think there is a mistake here:

u=d+w∗e(−vy/κ)
did you want to write this ##u(y)=c_{1}+c_{2}e^{-\frac{v}{\kappa}y}##, where ##c_{1},c_{2}## are "generic" constants ?

Ssnow
 
##v,\kappa## are constant ? What is ##w## ?

I think there is a mistake here:



did you want to write this ##u(y)=c_{1}+c_{2}e^{-\frac{v}{\kappa}y}##, where ##c_{1},c_{2}## are "generic" constants ?

Ssnow
Yes, sorry about that.
 

Ssnow

Gold Member
488
144
Ok, then ## u(y)=c_{1}+c_{2}e^{-\frac{v}{\kappa}y}## is the general solution...
Ssnow
 
Ok, then ## u(y)=c_{1}+c_{2}e^{-\frac{v}{\kappa}y}## is the general solution...
Ssnow
Yes, but I want to know how to get that solution from the differential equation.
 

Ssnow

Gold Member
488
144
In order to know "how" to find the general solution there is a general method for second order differential equations that consists to pass to the characteristic equation:

## \lambda^2+\frac{v}{\kappa} \lambda =0##

that has solutions ## \lambda_{1}=0## and ##\lambda_{2}=-\frac{v}{\kappa}##. When we have ##\Delta >0## the solution is a linear combination of exponentials:

##u(y)= c_{1}e^{0\cdot y}+c_{2}e^{-\frac{v}{\kappa}y}##

we write simply ##u(y)=c_{1}+c_{2}e^{-\frac{v}{\kappa}y}## (for details on this method I suggest any elementary book on differential equations)

Another way is to observe that:

## -\frac{v}{\kappa}y=\ln{\frac{d u}{dy}} +c ##

is equivalent to:

## -\frac{v}{\kappa}y-c=\ln{\frac{d u}{dy}} ##
that is

## e^{-\frac{v}{\kappa}y-c}=\frac{du}{dy}##

and the differential equation is: ##\frac{du}{dy}=Ke^{-\frac{v}{\kappa}y}## with ##K=e^{-c}## that is a constant.
This admit a simple solution that is the exponential itself, (you can think on this ...)
Ssnow
 

Ssnow

Gold Member
488
144
Precisely ,our equation is a second order homogeneous differential equation ...
Ssnow
 
In order to know "how" to find the general solution there is a general method for second order differential equations that consists to pass to the characteristic equation:

## \lambda^2+\frac{v}{\kappa} \lambda =0##

that has solutions ## \lambda_{1}=0## and ##\lambda_{2}=-\frac{v}{\kappa}##. When we have ##\Delta >0## the solution is a linear combination of exponentials:

##u(y)= c_{1}e^{0\cdot y}+c_{2}e^{-\frac{v}{\kappa}y}##

we write simply ##u(y)=c_{1}+c_{2}e^{-\frac{v}{\kappa}y}## (for details on this method I suggest any elementary book on differential equations)

Another way is to observe that:

## -\frac{v}{\kappa}y=\ln{\frac{d u}{dy}} +c ##

is equivalent to:

## -\frac{v}{\kappa}y-c=\ln{\frac{d u}{dy}} ##
that is

## e^{-\frac{v}{\kappa}y-c}=\frac{du}{dy}##

and the differential equation is: ##\frac{du}{dy}=Ke^{-\frac{v}{\kappa}y}## with ##K=e^{-c}## that is a constant.
This admit a simple solution that is the exponential itself, (you can think on this ...)
Ssnow
I understand everything that you showed me except the line:
## e^{-\frac{v}{\kappa}y-c}=\frac{du}{dy}##

How does that come from the original differential equation?
##-v(du/dy) = κ(d^2(u)/dy^2) ##
 

Ssnow

Gold Member
488
144
This is because ##\frac{1}{\frac{du}{dy}}\frac{d^{2}u}{dy^2}## is equal to ##\frac{d}{dy} \left(\ln{\frac{du}{dy}}\right)##. In fact:

##-v\frac{du}{dy}=\kappa\left(\frac{d^2u}{dy^2}\right)##

is equivalent to

##-\frac{v}{\kappa}=\frac{1}{\frac{du}{dy}}\left(\frac{d^2u}{dy^2}\right)##

that is equivalent to

##-\frac{v}{\kappa}=\frac{d}{dy} \left(\ln{\frac{du}{dy}}\right)##

now integrating both sides respect to ##y## we have : ##-\frac{v}{\kappa}y=\ln{\frac{du}{dy}}+c##.

Ssnow
 

Ray Vickson

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1. The problem statement, all variables and given/known data
integrate:

##-v(du/dy) = κ(d^2(u)/dy^2) ##

to obtain:

## (-v/κ)y = ln(du/dy) + c##

and finally:

##u = d + w*e^(-vy/κ)##

2. Relevant equations
##-v(du/dy) = κ(d^2(u)/dy^2) ##
## (-v/κ)y = ln(du/dy) + c##

3. The attempt at a solution
## (-v/κ)dy = d(u) ##

which gives:

## (-vy/κ) + C = u ##
Set ##w = du/dy## and then write the DE as ##dw/dy = r w,## where ##r = v/\kappa.## Thus, ##w = c e^{ry}## and ##u = \int w(y) \, dy.##
 
This is because ##\frac{1}{\frac{du}{dy}}\frac{d^{2}u}{dy^2}## is equal to ##\frac{d}{dy} \left(\ln{\frac{du}{dy}}\right)##. In fact:

##-v\frac{du}{dy}=\kappa\left(\frac{d^2u}{dy^2}\right)##

is equivalent to

##-\frac{v}{\kappa}=\frac{1}{\frac{du}{dy}}\left(\frac{d^2u}{dy^2}\right)##

that is equivalent to

##-\frac{v}{\kappa}=\frac{d}{dy} \left(\ln{\frac{du}{dy}}\right)##

now integrating both sides respect to ##y## we have : ##-\frac{v}{\kappa}y=\ln{\frac{du}{dy}}+c##.

Ssnow
Is there a proof for:
##\frac{1}{\frac{du}{dy}}\frac{d^{2}u}{dy^2}## = ##\frac{d}{dy} \left(\ln{\frac{du}{dy}}\right)##
 

Ssnow

Gold Member
488
144
Is there a proof for:
##\frac{1}{\frac{du}{dy}}\frac{d^{2}u}{dy^2}## = ##\frac{d}{dy} \left(\ln{\frac{du}{dy}}\right)##
The proof is an application of the derivative rule ## \frac{d}{dy}(f(g(y)))=\frac{df}{dy}(g(y))\cdot \frac{dg}{dy}(y)##, where ##f(y)=\ln(y)## and ##g(y)=\frac{du}{dy}(y)##.
Ssnow
 
The proof is an application of the derivative rule ## \frac{d}{dy}(f(g(y)))=\frac{df}{dy}(g(y))\cdot \frac{dg}{dy}(y)##, where ##f(y)=\ln(y)## and ##g(y)=\frac{du}{dy}(y)##.
Ssnow
Thanks!
 
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Since the problem asks you to integrate use a common method. Let p = du/dy to find a first order separable equation in p. Once you have p you then have u by integration. The method is much more general than this.
 

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