Question

b) If parametric equations of a flow line are x = x(t), y = y(t), explain why these functions satisfy the differential equations dx/dt = x and dy/dt = −y. Solve the differential equations to find an equation of the flow line that passes through the point (x, y) = (−1, −1). , where x 0

Answer 1

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!!!