Question

Solve the differential equation `dy/dt = 4 sqrt(yt)` y(1)=6?

Answer 1

`y=(4/3t^(3/2)+sqrt6-4/3)^2`

Explanation:

We should separate the variables here by treating `dy/dt` as a division problem to get all terms with `y` and all terms with `t` on the same side of the equation.

`dy/dt=4sqrtysqrtt`

`dy/sqrty=4sqrttdt`

Integrating both sides and rewriting with fractional exponents:

`inty^(-1/2)dy=4intt^(1/2)dt`

Using typical integration rules:

`y^(1/2)/(1/2)=4(t^(3/2)/(3/2))+C`

`2sqrty=8/3t^(3/2)+C`

Solving for `y` gives:

`y=(4/3t^(3/2)+C)^2`

We were given the initial condition `y(1)=6`, which we can use to solve for `C`:

`6=(4/3(1)^(3/2)+C)^2`

`sqrt6=4/3+C`

`C=sqrt6-4/3`

Then:

`y=(4/3t^(3/2)+sqrt6-4/3)^2`

Answer 2

`y = 16/9 ( t^(3/2) + (3 sqrt(6))/4 - 1)^2`

and:

`y = 16/9 ( t^(3/2) - (3 sqrt(6))/4 - 1)^2`

Explanation:

`dy/dt = 4 sqrt(yt)`

This is separable.

`1/sqrt y dy/dt = 4 sqrt t`

Differentiate both sides wrt t:

`int 1/sqrt y dy/dt \dt = 4 int sqrt t \ dt`

Chain rules allows us to re-write first term:

`int 1/sqrt y \ dy = 4 int sqrt t \ dt`

Then integrate:

`2 sqrt y = 4 2/3 t^(3/2) + C`

`implies sqrt y = 4/3 t^(3/2) + C`

`implies y = 16/9 ( t^(3/2) + C)^2`

Apply the IV: `y(1) = 6`

`implies 6 = 16/9 ( 1 + C)^2`

`implies C = pm sqrt(54/16) - 1 = pm (3 sqrt(6))/4 - 1`

So:

`implies y = 16/9 ( t^(3/2) + (3 sqrt(6))/4 - 1)^2`

And:

`implies y = 16/9 ( t^(3/2) - (3 sqrt(6))/4 - 1)^2`