Answer:
The equation of the the flow line that passes through the point (x, y) = (−1, −1) is
In y + In x = 0 or in another form, xy = 1.
Step-by-step explanation:
The pathline equation for a vector field is given by F(x,y) = xî - yj
The velocity vector field for the streamline of the flow is given by
V(x, y) = (dx/dt)î + (dy/dt)j
From the question, it is given that
(dx/dt) = x
(dy/dt) = -y
Hence, the velocity vector field for the streamline of the flow in question is
V(x, y) = xî - yj
which coincides with the pathline vector field of the flow.
The only time the pathline and streamline vector field coincide and have the same equation is when the flow is a steady state flow.
That is, the properties of the fluid flowing isn't changing with time!
Hence, this flow is a steady state flow!
We're told to solve the differential equation.
(dx/dt) = x
(dy/dt) = -y
but
(dy/dx) = (dy/dt) × (dt/dx)
(dy/dx) = -y/x
(dy/y) = -(dx/x)
∫(dy/y) = -∫ (dx/x)
In y = - In x + c
where c is the constant of integration
In y + In x = c
In (xy) = c
Inserting the values of (x, y) given in the question,
In (-1 × -1) = c
In 1 = c
0 = c
c = 0
In y + In x = 0
In (yx) = 0
xy = e⁰ = 1
xy = 1
So, the equation of the the flow line that passes through the point (x, y) = (−1, −1) is
In y + In x = 0 or in another form, xy = 1
Hope this Helps!!!
