Q.158. Differential equation.

Question 158.
Functions x(t) and y(t) satisfy dx/dt = -x^2y , dy/dt = -xy^2. when t = 0, x = 1 and y = 2.
Express dy/dx in terms of x and y and hence obtain y as a function of x and then deduce that dx/dt = -2x^3 and obtain x as a function of t for t> or equal to 0.

Answer 158.
dy/dx
= (dy/dt) / (dx/dt)
= ( - xy^2) / (- x^2y)
= y/x
=> dy/y = dx/x
=> ∫ dy/y = ∫ dx/x
=> ln lyl = ln lxl + c

Plugging the given x = 1 and y = 2,
=> c = ln2
=> ln lyl = ln lxl + ln2
=> y = 2x
=> dx/dt = - x^2y = - x^2 * (2x) = - 2x^3
=> (- 1/2) ∫ dx/x^3 = ∫ dt
=> (1/4) x^(-2) = t + c'

Plugging t = 0 and x = 1,
c' = 1/4
=> (1/4) x^(-2) = t + 1/4
=> 1/x^2 = 4t + 1
=> x = ± √[1/(4t + 1)].

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