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 differential equation is $(2 y-1) dx - (2 x+3) dy = 0$. Rearranging the terms,we get $(2 y-1) dx = (2 x+3) dy$. Separating the variables,we

Vedclass · Accepted Answer

$\frac{2 x+3}{2 y-1} = c$

Vedclass · Answer

(B) Given differential equation is $(2 y-1) dx - (2 x+3) dy = 0$. Rearranging the terms,we get $(2 y-1) dx = (2 x+3) dy$. Separating the variables,we have $\frac{dx}{2 x+3} = \frac{dy}{2 y-1}$. Integrating both sides,we get $\int \frac{dx}{2 x+3} = \int \frac{dy}{2 y-1}$. This results in $\frac{1}{2} \ln|2 x+3| = \frac{1}{2} \ln|2 y-1| + C_1$. Multiplying by $2$,we get $\ln|2 x+3| = \ln|2 y-1| + 2C_1$. Let $2C_1 = \ln|c|$,then $\ln|2 x+3| - \ln|2 y-1| = \ln|c|$. Using the property $\ln(a) - \ln(b) = \ln(\frac{a}{b})$,we get $\ln|\frac{2 x+3}{2 y-1}| = \ln|c|$. Taking the exponential of both sides,we obtain $\frac{2 x+3}{2 y-1} = c$.

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

Similar Questions

The general solution of the differential equation $y dx + (1 + x^2) \tan^{-1} x dy = 0$ is

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:

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

The particular solution of the differential equation $\cos \left(\frac{dy}{dx}\right) = 0.5$ with the condition $y = 1$ at $x = 0$ is:

The solution of the differential equation $e^x y dx + e^x dy + x dx = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module