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

Mar 7, 2017

$y = {\left(\frac{4}{3} {t}^{\frac{3}{2}} + \sqrt{6} - \frac{4}{3}\right)}^{2}$

#### Explanation:

We should separate the variables here by treating $\frac{\mathrm{dy}}{\mathrm{dt}}$ as a division problem to get all terms with $y$ and all terms with $t$ on the same side of the equation.

$\frac{\mathrm{dy}}{\mathrm{dt}} = 4 \sqrt{y} \sqrt{t}$

$\frac{\mathrm{dy}}{\sqrt{y}} = 4 \sqrt{t} \mathrm{dt}$

Integrating both sides and rewriting with fractional exponents:

$\int {y}^{- \frac{1}{2}} \mathrm{dy} = 4 \int {t}^{\frac{1}{2}} \mathrm{dt}$

Using typical integration rules:

${y}^{\frac{1}{2}} / \left(\frac{1}{2}\right) = 4 \left({t}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right) + C$

$2 \sqrt{y} = \frac{8}{3} {t}^{\frac{3}{2}} + C$

Solving for $y$ gives:

$y = {\left(\frac{4}{3} {t}^{\frac{3}{2}} + C\right)}^{2}$

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

$6 = {\left(\frac{4}{3} {\left(1\right)}^{\frac{3}{2}} + C\right)}^{2}$

$\sqrt{6} = \frac{4}{3} + C$

$C = \sqrt{6} - \frac{4}{3}$

Then:

$y = {\left(\frac{4}{3} {t}^{\frac{3}{2}} + \sqrt{6} - \frac{4}{3}\right)}^{2}$

Mar 7, 2017

$y = \frac{16}{9} {\left({t}^{\frac{3}{2}} + \frac{3 \sqrt{6}}{4} - 1\right)}^{2}$

and:

$y = \frac{16}{9} {\left({t}^{\frac{3}{2}} - \frac{3 \sqrt{6}}{4} - 1\right)}^{2}$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dt}} = 4 \sqrt{y t}$

This is separable.

$\frac{1}{\sqrt{y}} \frac{\mathrm{dy}}{\mathrm{dt}} = 4 \sqrt{t}$

Differentiate both sides wrt t:

$\int \frac{1}{\sqrt{y}} \frac{\mathrm{dy}}{\mathrm{dt}} \setminus \mathrm{dt} = 4 \int \sqrt{t} \setminus \mathrm{dt}$

Chain rules allows us to re-write first term:

$\int \frac{1}{\sqrt{y}} \setminus \mathrm{dy} = 4 \int \sqrt{t} \setminus \mathrm{dt}$

Then integrate:

$2 \sqrt{y} = 4 \frac{2}{3} {t}^{\frac{3}{2}} + C$

$\implies \sqrt{y} = \frac{4}{3} {t}^{\frac{3}{2}} + C$

$\implies y = \frac{16}{9} {\left({t}^{\frac{3}{2}} + C\right)}^{2}$

Apply the IV: $y \left(1\right) = 6$

$\implies 6 = \frac{16}{9} {\left(1 + C\right)}^{2}$

$\implies C = \pm \sqrt{\frac{54}{16}} - 1 = \pm \frac{3 \sqrt{6}}{4} - 1$

So:

$\implies y = \frac{16}{9} {\left({t}^{\frac{3}{2}} + \frac{3 \sqrt{6}}{4} - 1\right)}^{2}$

And:

$\implies y = \frac{16}{9} {\left({t}^{\frac{3}{2}} - \frac{3 \sqrt{6}}{4} - 1\right)}^{2}$