The general solution of the differential equa | vedclass

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

Question

(A) Given differential equation is $xy(y+2) dy + (y^3-1) dx = 0$. Rearranging the terms,we get $\frac{dx}{x} + \frac{y(y+2)}{y^3-1} dy = 0$. Integrati

Vedclass · Accepted Answer

$\log |x| + \frac{1}{3} \log |y^3-1| + \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{2y+1}{\sqrt{3}} ight) = c$

Vedclass · Answer

(A) Given differential equation is $xy(y+2) dy + (y^3-1) dx = 0$. Rearranging the terms,we get $\frac{dx}{x} + \frac{y(y+2)}{y^3-1} dy = 0$. Integrating both sides,$\int \frac{dx}{x} + \int \frac{y^2+2y}{y^3-1} dy = c$. Using partial fractions for $\frac{y^2+2y}{(y-1)(y^2+y+1)} = \frac{A}{y-1} + \frac{By+C}{y^2+y+1}$. Solving for constants,we find $A = 1$,$B = 0$,$C = 1$. So,$\int \frac{dx}{x} + \int \frac{1}{y-1} dy + \int \frac{1}{y^2+y+1} dy = c$. $\log |x| + \log |y-1| + \int \frac{1}{(y+1/2)^2 + (\sqrt{3}/2)^2} dy = c$. $\log |x(y-1)| + \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{2y+1}{\sqrt{3}} ight) = c$. Since $y^3-1 = (y-1)(y^2+y+1)$,the solution is $\log |x| + \frac{1}{3} \log |y^3-1| + \frac{2}{\sqrt{3}} 	an^{-1} \left( \frac{2y+1}{\sqrt{3}} ight) = c$.

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

Similar Questions

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

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

The solution of the differential equation $2x \frac{dy}{dx} - y = 0$,with the initial condition $y(1) = 2$,represents which of the following curves?

The general solution of the differential equation $(x-2y+1)dy-(3x-6y+2)dx=0$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module