Question

8ap^3=27y

Answer 1

Given: $8ap^3=27y$ , where $p=\frac{dy}{dx}$

Now $8ap^3=27y \Rightarrow p^3=\frac{27y}{8a} \Rightarrow p^3-\frac{27y}{8a}=0$

$\Rightarrow p^3-(\frac{3y^\frac{1}{3}}{2a^\frac{1}{3}})^3=0$

$\Rightarrow (p-\frac{3y^\frac{1}{3}}{2a^\frac{1}{3}})(p^2-\frac{3y^\frac{1}{3}}{2a^\frac{1}{3}}p+\frac{9y^\frac{2}{3}}{4a^\frac{2}{3}})=0$

$\Rightarrow p=\frac{3y^\frac{1}{3}}{2a^\frac{1}{3}}$ , $p=\frac{\frac{3y^\frac{1}{3}}{2a^\frac{1}{3}}\pm \sqrt{\frac{9y^\frac{2}{3}}{4a^\frac{2}{3}}-\frac{9y^\frac{2}{3}}{a^\frac{2}{3}}}}{2}$

$\Rightarrow p=\frac{3y^\frac{1}{3}}{2a^\frac{1}{3}},p=\frac{\frac{3y^\frac{1}{3}}{2a^\frac{1}{3}}\pm \frac{3\sqrt{3}y^\frac{1}{3}}{2a^\frac{1}{3}}i}{2}$

Observe that $p=\frac{\frac{3y^\frac{1}{3}}{2a^\frac{1}{3}}\pm \frac{3\sqrt{3}y^\frac{1}{3}}{2a^\frac{1}{3}}i}{2}$ are imaginary roots, so it can be discarded.

Now let ustake $p=\frac{3y^\frac{1}{3}}{2a^\frac{1}{3}}$

$\Rightarrow \frac{dy}{dx}=\frac{3y^\frac{1}{3}}{2a^\frac{1}{3}}$

Use separation of variables to solve the differential equation.

By using the separation of variables, we have

$\frac{dy}{y^\frac{1}{3}}=\frac{3}{2a^\frac{1}{3}}dx$

Integrating on both sides, we get

$\int \frac{dy}{y^\frac{1}{3}}=\frac{3}{2a^\frac{1}{3}}\int dx$

$\Rightarrow \frac{y^\frac{2}{3}}{2/3}=\frac{3}{2a^\frac{1}{3}}x+c$

$\Rightarrow y^\frac{2}{3}=xa^\frac{-1}{3}+\frac{2c}{3}$

$\Rightarrow y^\frac{2}{3}=xa^\frac{-1}{3}+C,C=\frac{2c}{3}$

Therefore, general solution is

$y-(xa^\frac{-1}{3}+C)^\frac{3}{2}=0$