Finding the value of the term named 'K'

zorro · #1 $Old$ November 1st, 2009, 03:27 AM

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

HallsofIvy · #2 $Old$ November 1st, 2009, 05:05 AM

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.

Prove It · #3 $Old$ November 1st, 2009, 05:11 AM

Quote:

Originally Posted by zorro $View Post$

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.