# Thread: Finding the value of the term named 'K'

1. ## 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

2. 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.

3. 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.

4. ## After going to the steps i am stuck at this eq

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

$-2e^t sint (sint + cost)^2 = K e^t (sint cost) - sint^2 - cost$

5. 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

6. ## Is y'' correct

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...

7. 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" 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

8. ## Need ur suggestion

Originally Posted by Coomast
"Mate" 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'

9. 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

10. ## Stuck again !!!

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

11. 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

12. ## Thanks mate ........

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