The particular solution of the differential e | vedclass

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

Question

(B) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.Let $x+y = v$. Then $1 + \frac{dy}{dx} = \frac{dv}{dx}$,so $\frac{dy}{dx} =

Vedclass · Accepted Answer

$y - x + \frac{1}{3} = \log |x+y|$

Vedclass · Answer

(B) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.Let $x+y = v$. Then $1 + \frac{dy}{dx} = \frac{dv}{dx}$,so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.Substituting into the equation: $\frac{dv}{dx} - 1 = \frac{v+1}{v-1}$.$\frac{dv}{dx} = \frac{v+1}{v-1} + 1 = \frac{v+1+v-1}{v-1} = \frac{2v}{v-1}$.Separating variables: $\int \frac{v-1}{2v} dv = \int dx$.$\frac{1}{2} \int (1 - \frac{1}{v}) dv = x + C$.$\frac{1}{2} (v - \log |v|) = x + C$.Substitute $v = x+y$: $\frac{x+y}{2} - \frac{1}{2} \log |x+y| = x + C$.Given $x = \frac{2}{3}$ and $y = \frac{1}{3}$,then $x+y = 1$.$\frac{1}{2} - \frac{1}{2} \log |1| = \frac{2}{3} + C$.$\frac{1}{2} - 0 = \frac{2}{3} + C \Rightarrow C = \frac{1}{2} - \frac{2}{3} = -\frac{1}{6}$.Substituting $C$ back: $\frac{x+y}{2} - \frac{1}{2} \log |x+y| = x - \frac{1}{6}$.Multiply by $2$: $(x+y) - \log |x+y| = 2x - \frac{1}{3}$.Rearranging: $y - x + \frac{1}{3} = \log |x+y|$.

The particular solution of the differential equation $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$ when $x = \frac{2}{3}$ and $y = \frac{1}{3}$ is

Similar Questions

The solution of the differential equation $e^{\frac{dy}{dx}} = x+1$ with the initial condition $y(0) = 5$ for $x \in (-1, \infty)$ is:

Given that the slope of the tangent to a curve $y=y(x)$ at any point $(x, y)$ is $\frac{2y}{x^2}$. If the curve passes through the centre of the circle $x^2+y^2-2x-2y=0$,then its equation is

The solution of the differential equation $x \cos y \, dy = (x e^x \log x + e^x) \, dx$ is

If $\frac{dy}{dx} = y + 5$ and $y(0) = 4$, then $y(\log 2)$ is equal to:

The general solution of the differential equation $\frac{dy}{dx} = \frac{x+y-3}{x+y-7}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module