The solution of the differential equation,y,d | vedclass

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

Question

(B) Given the differential equation: $y\,dx + (x + x^2y)dy = 0$Rearranging the terms,we get: $y\,dx + x\,dy + x^2y\,dy = 0$$(y\,dx + x\,dy) = -x^2y\,d

Vedclass · Accepted Answer

$-\frac{1}{xy} + \log y = c$

Vedclass · Answer

(B) Given the differential equation: $y\,dx + (x + x^2y)dy = 0$Rearranging the terms,we get: $y\,dx + x\,dy + x^2y\,dy = 0$$(y\,dx + x\,dy) = -x^2y\,dy$Dividing both sides by $x^2y^2$ (assuming $x, y 
eq 0$):$\frac{y\,dx + x\,dy}{x^2y^2} = -\frac{dy}{y}$Recognizing that $d(xy) = y\,dx + x\,dy$,the equation becomes:$\frac{d(xy)}{(xy)^2} = -\frac{dy}{y}$Integrating both sides:$\int (xy)^{-2} d(xy) = -\int \frac{1}{y} dy$$-\frac{1}{xy} = -\log|y| + c$Rearranging the terms,we get:$-\frac{1}{xy} + \log|y| = c$

The solution of the differential equation,$y\,dx + (x + x^2y)dy = 0$ is

Similar Questions

If $(2+\sin x) \frac{dy}{dx}+(y+1) \cos x=0$ and $y(0)=1$,then $y\left(\frac{\pi}{2}\right)$ is equal to

The general solution of $\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y$ is

Solve the differential equation: $\frac{dy}{dx} = e^{x+y}$

The solution of the differential equation $y(1+\log x) \frac{dx}{dy} - x \log x = 0$ is

The solution of $(x\sqrt{1 + y^2})dx + (y\sqrt{1 + x^2})dy = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module