Question

show that, y^2 + 2e^(-ky) +6
given y" at point (0,2) is -3/2

Answer 1

i dont understand question can u clarify

Answer 2

show what ?

Answer 3

or u need to find k, i think..

Answer 4

opps, sorry, the equation shuold be y^2 + 2e^(-ky) = 6

Answer 5

still, it isn't clear....
you have an equation, y^2 + 2e^(-ky) = 6
and given y" at point (0,2) is -3/2
what you need to show or find ?

Answer 6

sorry for the trouble. the qn i got is blur... The qn is show that, for y^2 + 2e^(-xy) = 6, the value of y" at the point (0,2) is - 3/2

Answer 7

$$y^2+2e^{-xy}=6\\\frac{d}{dx}\left[y^2+2e^{-xy}\right]=\frac{d}{dx}[6]\\2y\frac{dy}{dx}-2\left(y+x\frac{dy}{dx}\right)e^{-xy}=0\\2\frac{dy}{dx}\left(y-xe^{-xy}\right)=2ye^{-xy}\\\frac{dy}{dx}=\frac{ye^{-xy}}{y-xe^{-xy}}=\frac{y}{ye^{xy}-x}\\\quad\implies\frac{dy}{dx}=\frac2{2e^0-0}=1\\\frac{d^2y}{dx^2}=\frac{ye^{xy}-x-y(ye^{xy}(y+x\frac{dy}{dx})+\frac{dy}{dx}e^{xy}-1)}{(ye^{xy}-x)^2}\\\quad\implies\frac{d^2y}{dx^2}=\frac{2e^0-0-2(2e^0(2+0)+e^0-1)}{(2e^0-0)^2}=\frac{2-2(4)}{4}=\frac{2(-3)}{4}=-\frac32$$

Answer 8

wow thats some crazy latex work

Answer 9

Thanks a lot! =D