Chapter 7.4, Problem 1E

To determine

To find the initial value problem using separation of variable.

Answer to Problem 1E

  $\text{The required answer is, }y=\sqrt{x^2+3}$

Explanation of Solution

Given information: $\text{The equation is},\frac{dy}{dx}=\frac{x}{y}\text{ and }y=2\text{ when }x=1$

Concept used: Separable Differential Equation.

  $\begin{array}{l}\text{A differential equation of the form }dy/dx=f\left(y\right)g\left(x\right)\text{ is called separable}\text{.}\\ \text{We separate the variables by writing it in the form,}\\ \frac{1}{f\left(y\right)}dy=g\left(x\right)dx.\\ \text{The solution is found by antidifferentiating each side with respect to its thusly}\\ \text{isolated variable}\text{.}\end{array}$

Calculation : The required solution is obtained as,

  $\begin{array}{l}\text{The equation is separable because it can be written in the form of }dy/dx=x/y\text{, where}\\ f\left(y\right)=y\text{ and }g\left(x\right)=x.\text{ We separate the variables and antidifferentiate as follows}\text{. }\\ ydy=xdx\\ {\displaystyle \int yd}y={\displaystyle \int xdx}\\ \frac{1}{2}{y}^2=\frac{1}{2}{x}^2+C\\ y=\sqrt{x^2+C}\\ \text{we then apply the initial condition to find C}\text{.}\\ y=\sqrt{x^2+C}\\ \text{ 2}=\sqrt{1+C}\\ \text{ 4}=1+C\\ \text{ C}=3\\ y=\sqrt{x^2+3}\\ \text{Hence, the required answer is, }y=\sqrt{x^2+3}\end{array}$