Question : if $\displaystyle

y = e^t cos t , x = e^t sin t

$ and $\displaystyle y'' (x + y)^2 = K (xy' - y)$. Find K

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- Nov 1st 2009, 01:27 AMzorroFinding the value of the term named 'K'
Question : if $\displaystyle

y = e^t cos t , x = e^t sin t

$ and $\displaystyle y'' (x + y)^2 = K (xy' - y)$. Find K - Nov 1st 2009, 03:05 AMHallsofIvy
What have you

**tried**? Since $\displaystyle x= e^tcos(t)$ and $\displaystyle y= e^tsin(t)$, $\displaystyle y'= e^t (cos(t)+ sin(t)$ and $\displaystyle y"= e^t(cos(t)+ sin(t))+ e^t(-sin(t)+ cos(t))$$\displaystyle = 2e^tcos(t)$

Also $\displaystyle (x+y)^2= (e^t(cos(t)+ sin(t))^2= e^{2t}(cos^2(t)+ 2sin(t)cos(t)+ sin^2(t))$$\displaystyle = e^{2t}(1+ 2sin(t)cos(t))$.

Put those into the equations and see what you get. - Nov 1st 2009, 03:11 AMProve It
$\displaystyle x = e^t\sin{t}$.

$\displaystyle y = e^t\cos{t}$.

$\displaystyle y' = e^t\cos{t} - e^t\sin{t}$

$\displaystyle = e^t(\cos{t} - \sin{t})$.

$\displaystyle y'' = e^t(\cos{t} - \sin{t}) + e^t(-\sin{t} - \cos{t})$

$\displaystyle = -2e^t\sin{t}$.

Substitute this all into the DE.

$\displaystyle y'' (x + y)^2 = K (xy' - y)$

$\displaystyle -2e^t\sin{t}(e^t\sin{t} + e^t\cos{t})^2 = K[e^t\sin{t}e^t(\cos{t} - \sin{t}) - e^t\cos{t}]$

Now try to solve for K. - Nov 26th 2009, 02:22 PMzorroAfter going to the steps i am stuck at this eq
- Nov 28th 2009, 09:49 PMCoomast
I think this is not correct.

$\displaystyle y'=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt }}$

so, we have now:

$\displaystyle \frac{dy}{dt}=e^t\cdot \left(cos(t)-sin(t)\right)$

$\displaystyle \frac{dx}{dt}=e^t\cdot \left(cos(t)+sin(t)\right)$

$\displaystyle \frac{dy}{dx}=\frac{cos(t)-sin(t)}{cos(t)+sin(t)}$

from here you can calculate:

$\displaystyle y''=\frac{d^2y}{dx^2}=\frac{dt}{dx}\frac{d}{dt} \left(\frac{dy}{dt}\right)= \frac{\frac{d}{dt}\left(\frac{cos(t)-sin(t)}{cos(t)+sin(t)}\right) }{e^t\cdot (sin(t)+cos(t))}=...$

Putting all this in the given differential equation, you can obtain K.

(I got 2)

Coomast - Dec 1st 2009, 10:21 AMzorroIs y'' correct
- Dec 1st 2009, 10:45 AMCoomast
- Dec 1st 2009, 10:53 AMzorroNeed ur suggestion
- Dec 1st 2009, 10:59 AMCoomast
- Dec 1st 2009, 11:09 AMzorroStuck again !!!
- Dec 1st 2009, 11:15 AMCoomast
No, this is wrong. Expand the given factor of the DE as:

$\displaystyle xy'-y=x\cdot \frac{dy}{dx}-y=e^tsin(t) \cdot \left( \frac{cost(t)-sin(t)}{cost(t)+sin(t)} \right) -e^tcos(t)$

Can you proceed from here?

Coomast

edit: a "+" sign had to be a "-" sign....corrected - Dec 1st 2009, 11:27 AMzorroThanks mate ........
- Dec 1st 2009, 11:29 AMCoomast
Zorro,

You're welcome

Coomast