How to solve this diff. eqn

[Mr. Who]

 

please i need your help to solving this problem :

 

[Klaynos]

I think you can do it by seperarating the variables.

[Bignose]

I would be highly impressed if someone finds the analytic solution to this. I am not saying that there isn't one, but personally, if I had to have a solution to that non-linear second order beast, I would use a computer to give me the discrete approximate solution.

[CPL.Luke]

the solution is y=ct

 

just use integration by parts

 

^where c is solved for by using the initial condition

[Bignose]

the solution is y=ct

 

just use integration by parts

 

^where c is solved for by using the initial condition

 

???

 

Are we looking at the same equation here, guys?

 

if y=ct, then y'=c and y''=0. Plugging all that back in, you get

 

(ct)*0 + ©^2 = c

 

c has to equal 0 or 1 (not determined by the initial conditions). Not only that, but shouldn't there be two constants of integration, because of the second derivative?

[fredrik]

Probably homework? but anyway... here are some hints, instead of looking for y=f(t), one can look for t = f(y)?

 

[math]

yy'' +(y')^2 = y'

[/math]

 

Subst. w := yy' -y and the equation transforms into w' = 0. Thus w(t) = a

 

[math]

yy' -y = a

[/math]

 

Then against subst. z := y + a and you'll get

 

[math]

\left\{1-\frac{a}{z}\right\}dz = dt

[/math]

 

And you should find a solution

 

[math]

t = y+b-aln(y+a)

[/math]

 

a and b are constants. You might need to doublecheck the steps which I omitted beacuse I could be wrong.

 

/Fredrik

[ajb]

I asked mathematica and it gave me a horrible answer involving the productlog. As the productlog is an inverse function some of the solutions may be missing. anyway, here is my answer

 

[math]y(t) = -C2 -C2 \; Productlog \left( \frac{e^{-1 - \frac{t-C1}{C2}}}{C2}\right)[/math]

 

As to how to get to that answer I don't know. Best of luck working it out!

[fredrik]

Interesting. I think the productlog isn't counted as an elementary function though, but symbolic notation is always readable, but I did a quick check at that solution seems consistent with my suggested inverse.

 

The product log is defined as the inverse to [math]xe^x[/math] and if you use that relation it can be quickly seen to be consistent with the expression I found which is the inverse of ajb's expression.

 

[math]

y(t) = -C2 -C2 \; Productlog \left( \frac{e^{-1 - \frac{t-C1}{C2}}}{C2}\right)

[/math]

 

is by definition of the productlog the solution to

 

[math]

\frac{e^{-1 - \frac{t-C1}{C2}}}{C2} = - \frac{y(t) + C2}{C2} e^{-\frac{y(t) + C2}{C2}}

[/math]

 

Taking the logarithm of both sides

 

[math]

-1 - \frac{t-C1}{C2}= - ln \left\{ y(t) + C2 \right\} -\frac{y(t) + C2}{C2}

[/math]

 

and with redefining constants this seems consistent with the inverse form you can easily derive, in case the excerise requires proof of the productlog version.

 

/Fredrik

[ajb]

Thats great fredrik. I thought our solutions would be the same, but I did not check.

[CPL.Luke]

my bad I forgot one of the constants of integration

 

 

if you rearange the equation so that yy''=y'(1-y')

 

you can integrate to get

 

int(yy'')=y(1-y')-int(-yy'')+c1

 

c1=y(1-y')

 

which is seperable

 

int (y/(y-c1))=t

 

y(ln(y-c1))-ln(y-c1)+t+c2=t

 

y(lny(y-c1))-ln(y-c1)=c2

 

if c1 does not equal 0

 

then y=t+c2

[CPL.Luke]

woops there's a typor in the above equation

 

the final solution should be y(ln(y-c1)-t*ln(y-c1)=c_2

 

which can be solved for t

 

c_2/(ln(y-c1))+y=t

 

and the closest thing that I can get to for a seperated y solution is

 

c2(y-c1)^(y-1)=e^t

[Mr. Who]

hi every one,thank u very much

 

i am a student at math depatment ,it was my exam last question , i didn`t do any thing of that .

i try to find the solution by dividing by (y) , then adding -(y`/y) to the two parts of equation,

& i had no solution ,