# Finding the value of the term named 'K'

• Nov 1st 2009, 01:27 AM
zorro
Finding the value of the term named 'K'
Question : if $
y = e^t cos t , x = e^t sin t
$
and $y'' (x + y)^2 = K (xy' - y)$. Find K
• Nov 1st 2009, 03:05 AM
HallsofIvy
What have you tried? Since $x= e^tcos(t)$ and $y= e^tsin(t)$, $y'= e^t (cos(t)+ sin(t)$ and $y"= e^t(cos(t)+ sin(t))+ e^t(-sin(t)+ cos(t))$ $= 2e^tcos(t)$

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

Put those into the equations and see what you get.
• Nov 1st 2009, 03:11 AM
Prove It
Quote:

Originally Posted by zorro
Question : if $
y = e^t cos t , x = e^t sin t
$
and $y'' (x + y)^2 = K (xy' - y)$. Find K

$x = e^t\sin{t}$.

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

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

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

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

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

Substitute this all into the DE.

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

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

Now try to solve for K.
• Nov 26th 2009, 02:22 PM
zorro
After going to the steps i am stuck at this eq
Quote:

Originally Posted by Prove It
$x = e^t\sin{t}$.

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

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

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

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

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

Substitute this all into the DE.

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

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

Now try to solve for K.

After going thrght the steps and procedure i am stuck at this steps

$-2e^t sint (sint + cost)^2 = K e^t (sint cost) - sint^2 - cost$
• Nov 28th 2009, 09:49 PM
Coomast
Quote:

Originally Posted by HallsofIvy
What have you tried? Since $x= e^tcos(t)$ and $y= e^tsin(t)$, $y'= e^t (cos(t)+ sin(t)$ and $y"= e^t(cos(t)+ sin(t))+ e^t(-sin(t)+ cos(t))$ $= 2e^tcos(t)$

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

Put those into the equations and see what you get.

I think this is not correct.

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

so, we have now:

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

from here you can calculate:

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

Putting all this in the given differential equation, you can obtain K.
(I got 2)

Coomast
• Dec 1st 2009, 10:21 AM
zorro
Is y'' correct
Quote:

Originally Posted by Coomast
I think this is not correct.

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

so, we have now:

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

from here you can calculate:

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

Putting all this in the given differential equation, you can obtain K.
(I got 2)

Coomast

I am getting y'' as
$y'' = \frac{\frac{-2(cos^2 t + sin^2 t)}{(cos t + sin t)^2}}{e^t(cos t + sin t)}$

is that correct mate...
• Dec 1st 2009, 10:45 AM
Coomast
Quote:

Originally Posted by zorro
I am getting y'' as
$y'' = \frac{\frac{-2(cos^2 t + sin^2 t)}{(cos t + sin t)^2}}{e^t(cos t + sin t)}$

is that correct mate...

"Mate" (Happy) that's correct. Written more simply as:

$y''=\frac{-2}{e^t \cdot \left(cos(t)+sin(t)\right)^3}$

And putting all in the original DE gives for K....

Coomast
• Dec 1st 2009, 10:53 AM
zorro
Need ur suggestion
Quote:

Originally Posted by Coomast
"Mate" (Happy) that's correct. Written more simply as:

$y''=\frac{-2}{e^t \cdot \left(cos(t)+sin(t)\right)^3}$

And putting all in the original DE gives for K....

Coomast

mate this might be silly question but need to make sure
In the above equation where did $(cos^2 t + sin^2 t)$ go how did u get rid of it, did u take it as '1'
• Dec 1st 2009, 10:59 AM
Coomast
Quote:

Originally Posted by zorro
mate this might be silly question but need to make sure
In the above equation where did $(cos^2 t + sin^2 t)$ go how did u get rid of it, did u take it as '1'

Yes, Zorro, indeed, it is equal to 1. It is one of the most basic goniometric relations. Why would you assume this is not so?

coomast
• Dec 1st 2009, 11:09 AM
zorro
Stuck again !!!
Quote:

Originally Posted by Coomast
Yes, Zorro, indeed, it is equal to 1. It is one of the most basic goniometric relations. Why would you assume this is not so?

coomast

I guess mate i am doubting myself now days....
1 more question ....
$(xy' - y) = (e^t sint) (e^t (cos t - sin t)) - (e^t cos t)$

$= (e^t)^2 sin t cos t - (e^t)^2 sin^2 t - (e^t cos t)$

I am stuck her now
• Dec 1st 2009, 11:15 AM
Coomast
Quote:

Originally Posted by zorro
I guess mate i am doubting myself now days....
1 more question ....
$(xy' - y) = (e^t sint) (e^t (cos t - sin t)) - (e^t cos t)$

$= (e^t)^2 sin t cos t - (e^t)^2 sin^2 t - (e^t cos t)$

I am stuck her now

No, this is wrong. Expand the given factor of the DE as:

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

Can you proceed from here?

Coomast

edit: a "+" sign had to be a "-" sign....corrected
• Dec 1st 2009, 11:27 AM
zorro
Thanks mate ........
Quote:

Originally Posted by Coomast
No, this is wrong. Expand the given factor of the DE as:

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

Can you proceed from here?

Coomast