calculus help?

Question

find dy/dx in terms of x and y if x(e^y)-(x^2)y=(x^2)+1

Answer 1

For the terms on the left you need to apply product rule.

x*e^y *y' + e^y - ( x^2y' + 2xy ) = 2x

x*e^y *y' + e^y - x^2y' - 2xy = 2x now group all terms with y' on one side

x*e^y *y' - x^2y' = 2x +2xy - e^y

(x*e^y - x^2)y' = 2x +2xy - e^y so now

y' = (2x +2xy - e^y)/ (x*e^y - x^2)

Answer 2

Hello

dx(e^y) + e^y x dy -2xydx - x^2 dy = 2x dx

now put all the dx an one side and the dy on the other giving us

(e^y - 2xy -2x) dx = (x^2 -e^yx) dy so we have

(e^y -2xy - 2x)/(x^2 - e^yx) = dy/dx

Hope This Helps!!

Answer 3

This problem is called uv method

(if u are beginner in calculus),

x(e^y)-(x^2)y = x^2 + 1 ...............(eqn 1)

Differentiating eqn. (1) on both sides w.r.to x, we get

d/dx [x(e^y)] - d/dx[ (x^2)y] = d/dx[x^2 + 1] .....eqn (2)

first let us find d/dx[xe^y]

here let us take u = x, v=e^y

uv method: dy/dx = uv' + vu'
where u' = du/dx, v' =dv/dx

now d/dx[x(e^y)] = xe^y(y') + e^y(1)
where y' = dy/dx

next let us find d/dx[(x^2)y]. again by uv method,

d/dx[ (x^2)y] = (x^2)y' + y(2x)

and lastly

d/dx[x^2 + 1] = 2x + 0 = 2x

Substituting these in eqn (2), we get

xe^y(y') + e^y(1) - (x^2)y' + y(2x) = 2x

Separating y' term on one side, we get

y' [xe^y - x^2] = 2x - e^y + 2xy

y' = (2x - e^y + 2xy) / (xe^y - x^2)

Answer 4

use a TI-89 or pray or something