The general solution of the differential equa | vedclass

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

Question

(B) Given the differential equation: $x dy - y dx = 0$ Rearranging the terms,we get: $x dy = y dx$ Dividing both sides by $xy$ (assuming $x, y 
eq 0$

Vedclass · Accepted Answer

a straight line passing through the origin

Vedclass · Answer

(B) Given the differential equation: $x dy - y dx = 0$ Rearranging the terms,we get: $x dy = y dx$ Dividing both sides by $xy$ (assuming $x, y 
eq 0$): $\frac{dy}{y} = \frac{dx}{x}$ Integrating both sides: $\int \frac{dy}{y} = \int \frac{dx}{x}$ This yields: $\ln|y| = \ln|x| + C$ Using the property of logarithms,we can write the constant $C$ as $\ln|c|$: $\ln|y| = \ln|x| + \ln|c|$ $\ln|y| = \ln|cx|$ Taking the exponential of both sides: $y = cx$ This equation represents a family of straight lines passing through the origin.

The general solution of the differential equation $x dy - y dx = 0$ represents

Similar Questions

General solution of the differential equation $\cos x(1+\cos y) dx - \sin y(1+\sin x) dy = 0$ is

The solution of the differential equation $e^{2y} (1 + \ln x)dx + \csc y (2 + \cot y)dy = 0$ satisfying $y(1) = \frac{\pi}{2}$ is

The general solution of the differential equation $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}\right) = \sin \left(\frac{x-y}{2}\right)$ is

If $z = z(x)$ and $(2 + \cos x)\frac{dz}{dx} + (\sin x)z = \sin x$,$z(x) > 0$ and $z(\frac{\pi}{2}) = 3$,then $z(\frac{\pi}{3})$ equals -

If $y=y(x)$ satisfies the differential equation $8 \sqrt{x}(\sqrt{9+\sqrt{x}}) dy = (\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1} dx$ for $x>0$ and $y(0)=\sqrt{7}$,then find $y(256)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module