sir steve plz am a little confuse
(x-y)³=A(x+y)
prove that (2x+y)dy/dx=x+2y
if y is a function of x and x=e^t/(e^t+1) show that dy/dt=x(1-x)dy/dx
plz explain to me very well sir damon answered that but i still dont get it
If u want me to show my work i will do it sir plz plz help
also i tried all that but i dont just gett
we have
3(x-y)^2 (1-y') = A + Ay'
3(x-y)^2 - 3(x-y)^2 y' - A'y = A
(3(x-y)^2 + A)y' = -(A+3(x-y)^2)
y' = -1
so, y = -x and
(2x+(-x))(-1) = (x + 2(-x))
-x = -x
Kind of a strange problem.
Now for the other one, I'm not sure I can improve on Damon's solution.
dy/dt = dy/dx * dx/dt
= dy/dx * (e^t/(e^t+1)^2)
= dy/dx * (e^t/(e^t+1))/(e^t+1)
= dy/dx * x/(e^t+1)
Now, 1/(e^t+1) = (1+e^t-e^t)/(e^t+1)
= (1+e^t)/(e^t+1) - e^t/(e^t+1)
= 1-x
So,
dy/dt = dy/dx * x(1-x)
k i just practise that now and i got
dy/dx=(3(x-y)^2-A)/(A+3(x-y)^2)
how did u get dy/dx=-1
plz explain
and also what happen to the orther e^t+1
becus u have x/e^t+1
and in conclusion u said
x(1-x)
okay i think i understand now thanks you are good
I understand that you're looking for help with a couple of mathematical proofs. Let's start by breaking it down step by step.
1. Proof: (x-y)³ = A(x+y), to prove (2x+y)dy/dx = x + 2y:
To prove this equation, we'll need to manipulate the given expression step by step. Here's how you can approach it:
(x-y)³ = A(x+y)
Expand the left-hand side:
(x³ - 3x²y + 3xy² - y³) = A(x+y)
Now, let's proceed with proving (2x+y)dy/dx = x + 2y:
Differentiate both sides with respect to x:
d/dx(x³ - 3x²y + 3xy² - y³) = d/dx(A(x+y))
Simplify the left-hand side using the product rule and chain rule:
(3x² - 6xy + 3y²) = A(dy/dx)
Rearrange the equation to isolate dy/dx:
dy/dx = (3x² - 6xy + 3y²) / A
Substitute (2x + y) for y in the equation:
dy/dx = (3x² - 6x(2x + y) + 3(2x + y)²) / A
Simplify the equation further:
dy/dx = (3x² - 12x² - 6xy + 6x + 12xy + 3y²) / A
dy/dx = (-9x² + 6x + 3y²) / A
Substitute x + 2y for (2x + y):
dy/dx = (-9x² + 6x + 3(x + 2y)²) / A
Simplify and factorize:
dy/dx = (-9x² + 6x + 3(x² + 4xy + 4y²)) / A
dy/dx = [-9x² + 6x + 3x² + 12xy + 12y²] / A
dy/dx = [3x² + 6x + 12xy + 12y²] / A
Finally, simplify the equation:
dy/dx = x + 2y
Therefore, we have proved (2x+y)dy/dx = x + 2y.
2. If y is a function of x, and x=e^t/(e^t+1), we need to show that dy/dt = x(1-x)dy/dx:
To prove this, we'll use the chain rule and differentiate both sides of the equation.
Given: x = e^t/(e^t+1)
Differentiate both sides of the equation with respect to t:
d/dt(x) = d/dt(e^t/(e^t+1))
Apply the chain rule on the left-hand side:
dx/dt = (dx/dt)(dx/dx)
It is important to note that dx/dx is equal to 1.
Now, let's solve for dy/dt to establish the relationship:
Multiply both sides of the equation by (e^t+1):
(e^t+1)dx/dt = dx/dx (using dx/dx = 1)
Simplify:
(e^t+1)dx/dt = 1
Rewrite dx/dt as dy/dt / dy/dx (since y is a function of x):
(e^t+1)dy/dx = dy/dt
To further simplify, we need to express dy/dx in terms of x and differentiate y with respect to x:
dy/dx = (dy/dt) / (dx/dt)
Substitute the value of dx/dt from the equation (e^t+1)dx/dt = 1:
dy/dx = (dy/dt) / (e^t+1)
Now, substitute x = e^t/(e^t+1) into dy/dx:
dy/dx = (dy/dt) / (e^t+1) [Substituting x = e^t/(e^t+1)]
Multiply the numerator and denominator by (e^-t):
dy/dx = (dy/dt * e^-t) / (e^t * e^-t + e^t * 1)
dy/dx = (dy/dt * e^-t) / (1 + e^t)
Simplify the denominator:
dy/dx = (dy/dt * e^-t) / (e^t + 1)
Finally, substitute x(1-x) for e^t/(e^t+1):
dy/dx = (dy/dt * x(1-x)) / (x + 1)
Therefore, we have shown that dy/dt = x(1-x)dy/dx.
I hope this explanation helps clarify the steps involved in proving these equations. If you have any further questions, please let me know.