Chapter 1, Problem 1EIC

In Calculus you will see the symbol $y^{\prime }$

. For Exercises 1-4, treat $y^{\prime }$

as a variable and solve the equation for $y^{\prime }$.

$\frac{2x}{25}+\frac{2y}{9}{y}^{\prime }=0$

To determine

To calculate: The value of $y\text{'}$ in the equation $\frac{2x}{25}+\frac{2y}{9}y\text{'}=0$.

Answer to Problem 1EIC

Solution:

The solution of equation $\frac{2x}{25}+\frac{2y}{9}y\text{'}=0$ is $y\text{'}=-\frac{9x}{25y}$.

Explanation of Solution

Given information:

The given equation is $\frac{2x}{25}+\frac{2y}{9}y\text{'}=0$

Calculation:

Consider the given equation, $\frac{2x}{25}+\frac{2y}{9}y\text{'}=0$.

To remove the fraction, Multiply the equation by $225$.

$\begin{array}{c}225\left(\frac{2x}{25}+\frac{2y}{9}y\text{'}\right)=0\\ 18x+50yy\text{'}=0\end{array}$

Subtract $18x$ on both sides in above equation,

$50yy\text{'}=-18x$

Divided the equation by $50y$.

$\begin{array}{l}y\text{'}=-\frac{18x}{50y}\\ y\text{'}=-\frac{9x}{25y}\end{array}$

Hence, the value of $y\text{'}$ is $-\frac{9x}{25y}$ respectively.