How to Integrate a Differential Equation to Find u?

[shreddinglicks]

Homework Statement

integrate:

$-v(du/dy) = κ(d^2(u)/dy^2)$

to obtain:

$(-v/κ)y = ln(du/dy) + c$

and finally:

$u = d + w*e^(-vy/κ)$

Homework Equations

$-v(du/dy) = κ(d^2(u)/dy^2)$
$(-v/κ)y = ln(du/dy) + c$

The Attempt at a Solution

$(-v/κ)dy = d(u)$

which gives:

$(-vy/κ) + C = u$

 

[Ssnow]

$v,\kappa$ 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

 

[shreddinglicks]

$v,\kappa$ 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]

Ok, then $u(y)=c_{1}+c_{2}e^{-\frac{v}{\kappa}y}$ is the general solution...
Ssnow

 

[shreddinglicks]

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]

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]

Precisely ,our equation is a second order homogeneous differential equation ...
Ssnow

 

[shreddinglicks]

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]

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]

Homework Statement

integrate:

$-v(du/dy) = κ(d^2(u)/dy^2)$

to obtain:

$(-v/κ)y = ln(du/dy) + c$

and finally:

$u = d + w*e^(-vy/κ)$

Homework Equations

$-v(du/dy) = κ(d^2(u)/dy^2)$
$(-v/κ)y = ln(du/dy) + c$

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 $u = \int w(y) \, dy.$

 

[shreddinglicks]

Set $w = du/dy$ and then write the DE as $dw/dy = r w,$ where $r = v/\kappa.$ Thus, $w = c e^{ry}$ and $u = \int w(y) \, dy.$

I see, that is useful.

 

[shreddinglicks]

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]

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

 

[shreddinglicks]

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!

 

[alan2]

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.