Question

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52

$\frac{dy}{dx}=\frac{2x}{y^2},y\left(0\right)=4$

Short answer

The solution of the differential equation and obtain the solution of the initial-value problem $\frac{dy}{dx}=\frac{2x}{y^3}asy\left(x\right)=\sqrt[3]{3x^2+64}$

Step-by-step solution

Step 1. Given information

The given initial value problem$\frac{dy}{dx}=\frac{2x}{y^2},y\left(0\right)=4................\left(1\right)$

Step 2. Use antidifferentiation and/or separation of variables to solve each of the initial-value 

Note that the differential equation involved in (1) is of the form$\frac{dy}{dx}=p\left(x\right)p\left(y\right)$in which$p\left(x\right)=2x$, and$q\left(y\right)=\frac{1}{y^2}$. So, the differential equation can be solved by applying variable separable method. Thus, the solution of the differential equation involved in the initial- value problem is given by

$$\begin{gathered} \int y^{2}dy=\int 2xdx \\ \frac{1}{3}{y}^3=x^2+C_1 \\ {y}^3=3x^2+C(3C_1=C) \\ y=\sqrt[3]{3x^2+C} \end{gathered}$$

Now, use the given initial condition $y\left(0\right)=4$, that is take $x=0,y=4$ in the above result and evaluate the constant C.

$$\begin{gathered} 4=\sqrt[3]{C} \\ C=64 \end{gathered}$$

Substitute this value of the constant C in the solution of the differential equation and obtain the solution of the initial-value problem$\frac{dy}{dx}=\frac{2x}{y^3}asy\left(x\right)=\sqrt[3]{3x^2+64}$