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Real double roots question
- Thread starter oasi
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CaptainBlack
Well-known member
- Jan 26, 2012
- 890
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
- Feb 13, 2012
- 1,704
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
- Feb 13, 2012
- 1,704
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
$\chi$ $\sigma$
$\displaystyle y^{\ ''}+ a(x)\ y^{\ '} + b(x)\ y=0$ (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
$\chi$ $\sigma$