1. ## derivative

If $\displaystyle y = e^t cost, x=e^t sint$, and $\displaystyle y''(x+y)^2 = K(xy'-y)$
Find K

2. $\displaystyle y = e^t \cos{t}, x=e^t \sin{t}$

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

Well, $\displaystyle y = e^t \cos{t}$, so $\displaystyle y' = e^t \cos{t} - e^t \sin{t}$,
so $\displaystyle y'' = e^t \cos{t} - e^t \sin{t} - e^t \sin{t} - e^t \cos{t} = -2e^t \sin{t}$

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

$\displaystyle xy' = e^t( \cos{t} - \sin{t}) \times e^t \sin{t} = e^{2t} \sin{t} (\cos{t} - \sin{t})$

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

$\displaystyle \Rightarrow K = \frac{y''(x + y)^2}{xy' - y}$

$\displaystyle \Rightarrow K = \frac{-2e^t \sin{t} \times (e^{2t} + 2e^t \sin{t} \cos{t})}{e^{2t} \sin{t} (\cos{t} - \sin{t}) - e^t \cos{t}}$

$\displaystyle \Rightarrow K = \frac{-2e^t \sin{t} \times (e^t + 2 \sin{t} \cos{t})}{e^t \sin{t} (\cos{t} - \sin{t}) - \cos{t}}$

Hmmm I can't see this going where we want.....

3. Originally Posted by nzmathman
$\displaystyle y = e^t \cos{t}, x=e^t \sin{t}$

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

Well, $\displaystyle y = e^t \cos{t}$, so $\displaystyle y' = e^t \cos{t} - e^t \sin{t}$,
so $\displaystyle y'' = e^t \cos{t} - e^t \sin{t} - e^t \sin{t} - e^t \cos{t} = -2e^t \sin{t}$

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

$\displaystyle xy' = e^t( \cos{t} - \sin{t}) \times e^t \sin{t} = e^{2t} \sin{t} (\cos{t} - \sin{t})$

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

$\displaystyle \Rightarrow K = \frac{y''(x + y)^2}{xy' - y}$

$\displaystyle \Rightarrow K = \frac{-2e^t \sin{t} \times (e^{2t} + 2e^t \sin{t} \cos{t})}{e^{2t} \sin{t} (\cos{t} - \sin{t}) - e^t \cos{t}}$

$\displaystyle \Rightarrow K = \frac{-2e^t \sin{t} \times (e^t + 2 \sin{t} \cos{t})}{e^t \sin{t} (\cos{t} - \sin{t}) - \cos{t}}$

Hmmm I can't see this going where we want.....
That's because in this case, y isn't a function of x, it's a function of t.

Remember that $\displaystyle y' = \frac{dy}{dx}$.

You need to use the chain rule $\displaystyle y' = \frac{dy}{dt}\times\frac{dt}{dx}$.

Use implicit differentiation to find $\displaystyle y''$.

4. Ah yes!

I would have another go, but it is getting late and I'm getting tired...

5. Originally Posted by varunnayudu
If $\displaystyle y=e^t cost, x=e^t sint$, and $\displaystyle y''(x+y)^2 = K(xy' - y).$Find $\displaystyle K$

6. ## Sure....

i have done this .....

$\displaystyle y' = \frac{dy}{dt} = e^t (-sint)$

therefore

$\displaystyle y'' = \frac{d^2 y }{dt} = e^t (-cost)$

7. Originally Posted by varunnayudu
i have done this .....

$\displaystyle y' = \frac{dy}{dt} = e^t (-sint)$

therefore

$\displaystyle y'' = \frac{d^2 y }{dt} = e^t (-cost)$
First you should carefully review the product rule for differentiating a product of two functions. Then read post #3 more carefully.

8. well, take a look on what the previous posters replied and also the comment.. you should be able to get the answer..