The Student Room Group

differentiation

Given that y=x+ke^x where k is a constant, show that:
(1-x)d^2y/dx^2 + x(dy/dx) - y = 0.

where do i begin on this question??
Reply 1
Original post by elvss567
Given that y=x+ke^x where k is a constant, show that:
(1-x)d^2y/dx^2 + x(dy/dx) - y = 0.

where do i begin on this question??


Well, if y = x + ke^x then what's dy/dx? What's d^2y/dx^2?

What's (1-x)d^2y/dx^2 + x(dy/dx) - y?
Reply 2
Original post by Zacken
Well, if y = x + ke^x then what's dy/dx? What's d^2y/dx^2?

What's (1-x)d^2y/dx^2 + x(dy/dx) - y?


for dy/dx = 1 +ke^x
d^2y/dx^2 = ke^x
what do i do then?
Reply 3
Original post by elvss567
for dy/dx = 1 +ke^x
d^2y/dx^2 = ke^x
what do i do then?


You haven't answered my other question, but anyway:

If dy/dx = 1 + ke^x and d^2y/dx^2 = ke^x then what's (1-x)d^2y/dx^2? What's x(dy/dx)? What's (1-x)d^2y/dx^2 + x(dy/dx) - y?
Reply 4
Original post by Zacken
You haven't answered my other question, but anyway:

If dy/dx = 1 + ke^x and d^2y/dx^2 = ke^x then what's (1-x)d^2y/dx^2? What's x(dy/dx)? What's (1-x)d^2y/dx^2 + x(dy/dx) - y?


ohh so i'm just substituting in the equation for dy/dx and d2x/dy2 and y .
then just simplifying it down?
Reply 5
Original post by elvss567
ohh so i'm just substituting in the equation for dy/dx and d2x/dy2 and y .
then just simplifying it down?


yes
Reply 6
Original post by Zacken
yes


That makes so much more sense! Thankyou!:smile:

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