Chapter 2, Problem 1RE

To determine

The solution of the differential equation $\frac{dy}{dt}=\frac{2t^5}{5y^2}$.

Answer to Problem 1RE

The solution of the differential equation $\frac{dy}{dt}=\frac{2t^5}{5y^2}$ is $\underset{\_}{\frac{1}{5}\sqrt[3]{25t^6+125C}}$.

Explanation of Solution

Definition used:

Separable differential equation:

A first order differential equation that can be written in the form $g\left(y\right)dy=f\left(t\right)dt$ is called the separable differential equation.

Calculation:

Given that, the differential equation is, $\frac{dy}{dt}=\frac{2t^5}{5y^2}$.

It can be written in the form of separable differential equation as, $5y^{2}dy=2t^{5}dt$.

Now integrate both sides and obtain the general solution as follows.

$\begin{array}{l}5y^{2}dy=2t^{5}dt\\ {\displaystyle \int 5y^{2}dy}={\displaystyle \int 2t^{5}dt}\\ 5\left(\frac{y^3}{3}\right)=2\left(\frac{t^6}{6}\right)+c\\ 5y^3=6\left(\frac{t^6}{6}\right)+C\left(\because C=3c\right)\end{array}$

Simplify the above equation.

$\begin{array}{c}{y}^3=\left(\frac{t^6}{5}\right)+\frac{C}{5}\\ y=\sqrt[3]{\frac{t^6}{5}+\frac{C}{5}}\\ =\frac{1}{5}\sqrt[3]{25t^6+125C}\end{array}$

Hence, the solution of the differential equation $\frac{dy}{dt}=\frac{2t^5}{5y^2}$ is $\underset{\_}{\frac{1}{5}\sqrt[3]{25t^6+125C}}$.