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