The general solution of the differential equa | vedclass

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

Question

(C) Given the differential equation $\frac{dy}{dx} = \frac{x+2y-1}{x+2y+1}$.Let $t = x+2y$. Then $\frac{dt}{dx} = 1 + 2\frac{dy}{dx}$,which implies $\

Vedclass · Accepted Answer

$6(-x+y)+4 \log |3x+6y-1| = K$

Vedclass · Answer

(C) Given the differential equation $\frac{dy}{dx} = \frac{x+2y-1}{x+2y+1}$.Let $t = x+2y$. Then $\frac{dt}{dx} = 1 + 2\frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2}\left(\frac{dt}{dx} - 1\right)$.Substituting this into the equation: $\frac{1}{2}\left(\frac{dt}{dx} - 1\right) = \frac{t-1}{t+1}$.$\frac{dt}{dx} - 1 = \frac{2t-2}{t+1} \Rightarrow \frac{dt}{dx} = \frac{2t-2}{t+1} + 1 = \frac{2t-2+t+1}{t+1} = \frac{3t-1}{t+1}$.Separating the variables: $\int \frac{t+1}{3t-1} dt = \int dx$.To integrate $\frac{t+1}{3t-1}$,we write it as $\frac{1}{3} \int \frac{3t+3}{3t-1} dt = \frac{1}{3} \int \frac{3t-1+4}{3t-1} dt = \frac{1}{3} \int (1 + \frac{4}{3t-1}) dt$.This gives $\frac{1}{3} (t + \frac{4}{3} \log |3t-1|) = x + C$.Multiplying by $9$: $3t + 4 \log |3t-1| = 9x + 9C$.Substituting $t = x+2y$: $3(x+2y) + 4 \log |3(x+2y)-1| = 9x + K$.$3x + 6y + 4 \log |3x+6y-1| = 9x + K$.$6y - 6x + 4 \log |3x+6y-1| = K$.$6(-x+y) + 4 \log |3x+6y-1| = K$.

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

Similar Questions

The solution of $\frac{dy}{dx} = \frac{x \log(x^2) + x}{\sin y + y \cos y}$ is:

The solution of the differential equation $(1+e^{-x})(1+y^2) \frac{dy}{dx} = y^2$ which passes through the point $(0,1)$ is

Find a particular solution of the differential equation $(x+1) \frac{dy}{dx} = 2e^{-y} - 1$,given that $y = 0$ when $x = 0$.

The general solution of the differential equation $(2 y-1) dx - (2 x+3) dy = 0$ is

The general solution of the differential equation $\frac{x dy - y dx}{y} = 0$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module