Differential Equations Help

Question

Suppose that y(t) solves the ordinary differential equation

dy/dt = (t*y^-.5)/3 y(0)=1. Find y(1)

I tried solving using separable equations but am having a hard time.

Answer 1

It was a good idea to solve this problem using separation of variables. The first step is to remember the meaning of negative and fractional exponents from algebra. Unless you're worried about significant figures like in a chemistry course, an exponent of -0.5 is the same as an exponent of -1/2. A negative exponent moves the expression to the denominator and the fraction 1/2 in the exponent means a square root is taken. So...

(t*y^-.5)/3 = t/(3√y)

Now to separate variables, move everything involving a y to one side of the differential equation and everything involving a t to the other side.

dy/dt = t/(3√y) becomes √y dy= (1/3)t dt

Returning the square root symbol to a fractional exponent,

y^(1/2) dy= (1/3)t dt

will help to integrate the expression:

∫y^(1/2) dy= ∫(1/3)t dt

by using the power rule for integration to yield:

(2/3) y^(3/2) = t^2 / 6 + C.

We can use the fact that y(0) = 1 to solve for the integration constant C by substituting 1 for y and 0 for t:

(2/3) 1^(3/2) = (0)^2 / 6 + C

2/3 = C

Now we can determine y(1) by solving for y when t = 1:

(2/3) y^(3/2) = 1^2 / 6 + 2/3

y^(3/2) = (1/6 + 2/3)(3/2) = (5/6)(3/2) = 5/4

y = (5/4)^(2/3)

Answer 2

dy/dt = (t*y^-.5)/3

3dy = (t*y^-.5) * dt

3* y^0.5 dy = t * dt

2 * y^1.5 = 0.5 * t ^ 2 + C

y = ((0.5 * t ^ 2 + C) / 2) ^ (2/3)

y(0) = 1

1 = (C / 2) ^ (2/3)

C = 2

y(1) = ((0.5 * 1 ^ 2 + 2) / 2) ^ (2/3) = (2.5 / 2) ^ (2/3)