The solution of frac{dy}{dx} = (frac{x}{y})^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the differential equation: $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$.By simplifying the right side, we get: $\frac{dy}{dx} = (\frac{y}{x})^{1/3

Vedclass · Accepted Answer

$y^{2/3} - x^{2/3} = c$

Vedclass · Answer

(B) Given the differential equation: $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$.By simplifying the right side, we get: $\frac{dy}{dx} = (\frac{y}{x})^{1/3} = \frac{y^{1/3}}{x^{1/3}}$.Separating the variables, we have: $y^{-1/3} dy = x^{-1/3} dx$.Integrating both sides:$\int y^{-1/3} dy = \int x^{-1/3} dx$.Using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:$\frac{y^{2/3}}{2/3} = \frac{x^{2/3}}{2/3} + C_1$.Multiplying by $\frac{2}{3}$:$y^{2/3} = x^{2/3} + \frac{2}{3}C_1$.Let $c = \frac{2}{3}C_1$, then $y^{2/3} - x^{2/3} = c$.

The solution of $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$ is

Similar Questions

In which phase of cell division do the nuclear envelope and nucleolus reappear?

Find out the value of $K_C$ for the following reaction from the value of $K_P$
$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$
[Given : $K_P = 8 \times 10^{12} \ atm$ at $500 \ K$ use $R = 0.08 \ L \ atm \ mol^{-1} \ K^{-1}$ ]

Intercalary meristem results in

What is the cell wall of Protista composed of?

Vedclass Test Series

Exam Paper Generator

Online Exam Module