The solution of the differential equation x + | vedclass

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

Question

(B) Given the differential equation $x + y\frac{dy}{dx} = 2y$.Rearranging the terms,we get $y\frac{dy}{dx} = 2y - x$,which implies $\frac{dy}{dx} = \f

Vedclass · Accepted Answer

$\log (y - x) = c + \frac{x}{y - x}$

Vedclass · Answer

(B) Given the differential equation $x + y\frac{dy}{dx} = 2y$.Rearranging the terms,we get $y\frac{dy}{dx} = 2y - x$,which implies $\frac{dy}{dx} = \frac{2y - x}{y}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{2vx - x}{vx} = \frac{2v - 1}{v}$.$x\frac{dv}{dx} = \frac{2v - 1}{v} - v = \frac{2v - 1 - v^2}{v} = -\frac{(v - 1)^2}{v}$.Separating the variables: $\frac{v}{(v - 1)^2} dv = -\frac{dx}{x}$.Using partial fractions: $\frac{v}{(v - 1)^2} = \frac{v - 1 + 1}{(v - 1)^2} = \frac{1}{v - 1} + \frac{1}{(v - 1)^2}$.Integrating both sides: $\int \left( \frac{1}{v - 1} + \frac{1}{(v - 1)^2} \right) dv = -\int \frac{dx}{x}$.$\log |v - 1| - \frac{1}{v - 1} = -\log |x| + c$.Since $v = \frac{y}{x}$,we have $v - 1 = \frac{y - x}{x}$.$\log |\frac{y - x}{x}| + \log |x| = \frac{1}{\frac{y - x}{x}} + c$.$\log |y - x| = \frac{x}{y - x} + c$.

The solution of the differential equation $x + y\frac{dy}{dx} = 2y$ is

Similar Questions

$3 P(A) = P(B) = \frac{5}{13}$ and $P(A \mid B) = \frac{3}{5}$,then $P(A \cup B) = $ . . . . . . .

$A$ curve passes through the point $\left( 1, \frac{\pi}{4} \right)$ and its slope at any point is given by $\frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$. Find the equation of the curve.

One ticket is selected at random from $50$ tickets numbered $\{00, 01, 02, \ldots, 49\}$. Then the probability that the sum of the digits on the selected ticket is $8$,given that the product of these digits is zero,is

The solution of $x dy - y dx = \sqrt{x^2 + y^2} dx$ when $y(\sqrt{3}) = 1$ is

$f(x, y) = \frac{1}{x + y}$ is a homogeneous function of degree

Vedclass Test Series

Exam Paper Generator

Online Exam Module