for the sake of one good student

[logistic_guy]

Solve.

$\displaystyle u'' + \frac{1}{x}u' - \frac{1}{x^2}u = 0$

 

[logistic_guy]

One good student may wonder why the solution to this differential equation is $\displaystyle u(x) = c_1x + \frac{c_2}{x}$.

For the sake of completeness and of this one good student, we will show one way to do it. Be ready and be in touch with us in the next posts.

 

[logistic_guy]

There are many ways to solve it. I like to do it in this way:

Let $\displaystyle x = e^t$ and $\displaystyle u(x) = v(t)$

$\displaystyle \frac{du}{dx} = \frac{dv}{dx} = \frac{dv}{dt}\frac{dt}{dx}$

We have $\displaystyle t = \ln x$, then

$\displaystyle \frac{du}{dx} = \frac{dv}{dt}\frac{1}{x}$

$\displaystyle \frac{d^2u}{dx^2} = \frac{d}{dx}\left(\frac{dv}{dt}\frac{1}{x}\right) = \frac{d}{dx}\left(\frac{dv}{dt}\right)\frac{1}{x} + \frac{dv}{dt}\frac{d}{dx}\left(\frac{1}{x}\right)$

$\displaystyle = \frac{d}{dt}\left(\frac{dv}{dt}\right)\frac{dt}{dx}\frac{1}{x} + \frac{dv}{dt}\left(-\frac{1}{x^2}\right) = \frac{d^2v}{dt^2}\frac{1}{x^2} - \frac{dv}{dt}\frac{1}{x^2}$

Substitute the result back to the original differential equation.

$\displaystyle \frac{d^2v}{dt^2}\frac{1}{x^2} - \frac{dv}{dt}\frac{1}{x^2} + \frac{dv}{dt}\frac{1}{x^2} - v\frac{1}{x^2} = 0$

Simplify.

$\displaystyle \frac{d^2v}{dt^2} - v = 0$

This is a very basic differential equation and the roots of its characteristic equation are:

$\displaystyle r = \frac{\pm\sqrt{-4(1)(-1)}}{2(1)} = \pm 1$

Those roots give a solution of:

$\displaystyle v(t) = c_1e^t + c_2e^{-t} = c_1e^t + \frac{c_2}{e^{t}}$

But we already know that $\displaystyle v(t) = u(x)$ and $\displaystyle e^t = x$, then the solution becomes:

$\displaystyle u(x) = c_1x + \frac{c_2}{x}$

 

[khansaheb]

There are many ways to solve it. I like to do it in this way:

Let $\displaystyle x = e^t$ and $\displaystyle u(x) = v(t)$

$\displaystyle \frac{du}{dx} = \frac{dv}{dx} = \frac{dv}{dt}\frac{dt}{dx}$

We have $\displaystyle t = \ln x$, then

$\displaystyle \frac{du}{dx} = \frac{dv}{dt}\frac{1}{x}$

$\displaystyle \frac{d^2u}{dx^2} = \frac{d}{dx}\left(\frac{dv}{dt}\frac{1}{x}\right) = \frac{d}{dx}\left(\frac{dv}{dt}\right)\frac{1}{x} + \frac{dv}{dt}\frac{d}{dx}\left(\frac{1}{x}\right)$

$\displaystyle = \frac{d}{dt}\left(\frac{dv}{dt}\right)\frac{dt}{dx}\frac{1}{x} + \frac{dv}{dt}\left(-\frac{1}{x^2}\right) = \frac{d^2v}{dt^2}\frac{1}{x^2} - \frac{dv}{dt}\frac{1}{x^2}$

Substitute the result back to the original differential equation.

$\displaystyle \frac{d^2v}{dt^2}\frac{1}{x^2} - \frac{dv}{dt}\frac{1}{x^2} + \frac{dv}{dt}\frac{1}{x^2} - v\frac{1}{x^2} = 0$

Simplify.

$\displaystyle \frac{d^2v}{dt^2} - v = 0$

This is a very basic differential equation and the roots of its characteristic equation are:

$\displaystyle r = \frac{\pm\sqrt{-4(1)(-1)}}{2(1)} = \pm 1$

Those roots give a solution of:

$\displaystyle v(t) = c_1e^t + c_2e^{-t} = c_1e^t + \frac{c_2}{e^{t}}$

But we already know that $\displaystyle v(t) = u(x)$ and $\displaystyle e^t = x$, then the solution becomes:

$\displaystyle u(x) = c_1x + \frac{c_2}{x}$

You'll need to show that your solution satisfies the original DE.

 

[logistic_guy]

You'll need to show that your solution satisfies the original DE.

Nice ideathat did not cross my mind!

$\displaystyle u'(x) = c_1 - \frac{c_2}{x^2}$

$\displaystyle u''(x) = \frac{2c_2}{x^3}$

Substitude back.

$\displaystyle \frac{2c_2}{x^3} + \frac{1}{x}\left(c_1 - \frac{c_2}{x^2}\right) - \frac{1}{x^2}\left(c_1x + \frac{c_2}{x}\right) = 0$


$\displaystyle \frac{2c_2}{x^3} + \left(\frac{c_1}{x} - \frac{c_2}{x^3}\right) - \left(\frac{c_1}{x} + \frac{c_2}{x^3}\right) = 0$


$\displaystyle \frac{2c_2}{x^3} + \frac{c_1}{x} - \frac{c_2}{x^3} - \frac{c_1}{x} - \frac{c_2}{x^3} = 0$


$\displaystyle \frac{2c_2}{x^3} + \frac{c_1}{x} - \frac{c_1}{x} - \frac{2c_2}{x^3} = 0$


$\displaystyle 0 = 0$



I was hoping that the solution would not be satisfied

 

[logistic_guy]

Substitude back.

I hope that Sir @lookagain won't see this. Otherwise, a lecture of $\displaystyle 400$ pages will be given!