Asked by Susan
Hello, I have the next exercise about chain rule in multivariable calculus:
I have to show that the differential equation:
y(dz/dx)-x(dz/dy) = (y-x)z
can be change to the equation:
(dw/dv) = 0
using the new variables:
u= (x^2)+(y^2)
v = (1/x)+(1/y)
w = ln(z)-(x+y)
For that I have to do some steps,in the first one I have to find the derivative of w respect to v
My problem is that I don't know what to do in order to express that derivative of w respect to v
Please could anyone help me giving me some steps o what to do?
I would really appreaciate the help
I have to show that the differential equation:
y(dz/dx)-x(dz/dy) = (y-x)z
can be change to the equation:
(dw/dv) = 0
using the new variables:
u= (x^2)+(y^2)
v = (1/x)+(1/y)
w = ln(z)-(x+y)
For that I have to do some steps,in the first one I have to find the derivative of w respect to v
My problem is that I don't know what to do in order to express that derivative of w respect to v
Please could anyone help me giving me some steps o what to do?
I would really appreaciate the help
Answers
Answered by
Steve
dw/dv = (dw/dx) / (dv/dx)
dw/dx = ∂w/∂x + ∂w/∂y dy/dx + ∂w/∂z dz/dx
and similarly for u,v
See where you get with that.
dw/dx = ∂w/∂x + ∂w/∂y dy/dx + ∂w/∂z dz/dx
and similarly for u,v
See where you get with that.
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