# Real double roots question

#### CaptainBlack

##### Well-known member
It would help if you posted the full question (if there is any more, or its exact wording), but in this context it means something like give a "derivation of" that is provide a logical set of arguments that find the given function as a solution of the ODE.

CB

#### chisigma

##### Well-known member
The second order ODE that has solutions You have indicated is...

$\displaystyle y^{\ ''} + a\ y^{\ '}+ \frac{a^{2}}{4}\ y=0$ (1)

A possible answer to Your question is in a procedure that can be very useful in more complicated situations. The general solution of (1) is...

$\displaystyle y(x)= c_{1}\ u(x) + c_{2}\ v(x)$ (2)

... where $u(*)$ and $v(*)$ are mutually linearly independent, $c_{1}$ and $c_{2}$ two constants. From (1) we derive...

$\displaystyle u^{\ ''} + a\ u^{\ '}+ \frac{a^{2}}{4}\ u=0$

$\displaystyle v^{\ ''} + a\ v^{\ '}+ \frac{a^{2}}{4}\ v=0$ (3)

... and multiplying the first of (3) by v, the second of (3) by u and taking the difference...

$\displaystyle (v\ u^{\ ''} - u\ v^{\ ''}) + a\ (v\ u^{\ '} -u\ v^{\ '})=0$ (4)

Setting $\displaystyle z=v\ u^{\ '} -u\ v^{\ '}$ the (4) becomes...

$z^{\ '}+ a\ z=0$ (5)

... the solution of which is $\displaystyle z=c_{1}\ e^{-a\ x}$. Combining all the result we can write...

$\displaystyle \frac{v\ u^{\ '} -u\ v^{\ '}}{v^{2}}= \frac{d}{dx} (\frac{u}{v})= \frac{c_{1}\ e^{-a\ x}}{v^{2}} \implies \frac{u}{v}= c_{2}+c_{1}\ \int \frac{e^{-a\ x}}{v^{2}}\ dx \implies u= c_{2}\ v + c_{1}\ v\ \int \frac{e^{-a\ x}}{v^{2}}\ dx$ (6)

Now we can verify that $\displaystyle v(x)= e^{- \frac{a\ x}{2}}$ is solution of (1) so that from (6) we obtain...

$\displaystyle u(x)= c_{1}\ e^{- \frac{a\ x}{2}}\ \int \frac{e^{-a\ x}}{e^{-a\ x}}\ dx = c_{1}\ x\ e^{- \frac{a\ x}{2}}$ (7)

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
A more general application of the procedure of my previous post is the following. Let's suppose to have a second order ODE...

$\displaystyle y^{\ ''}+ a(x)\ y^{\ '} + b(x)\ y=0$ (1)

Proceeding as in my previous post, setting $\displaystyle z= v\ u^{\ '} - u\ v^{\ '}$ we arrive at...

$\displaystyle z^{\ '}+ a(x)\ z=0$ (2)

... a solution of which is...

$\displaystyle z= e^{- \int a(x)\ dx}$ (3)

... and from (3) we derive first...

$\displaystyle \frac{d}{dx} (\frac{u}{v})= \frac{z}{v^{2}}$ (4)

... and then...

$\displaystyle u=v\ \int \frac{z}{v^{2}}\ dx$ (5)

The (5) is useful because it permits You, if a solution v(x) of (1) is known, to obtain another solution u(x) independent from it. A 'curiousity' is the fact that (5) is valid no matter which is b(x) in (1)...

Kind regards

$\chi$ $\sigma$