Find the general solution of the differential | vedclass

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

Question

(A) The given differential equation is: $e^{x} 	an y \, dx + (1 - e^{x}) \sec^{2} y \, dy = 0$. Rearranging the terms,we get: $(1 - e^{x}) \sec^{2} y

Vedclass · Accepted Answer

$	an y = C(1 - e^{x})$

Vedclass · Answer

(A) The given differential equation is: $e^{x} 	an y \, dx + (1 - e^{x}) \sec^{2} y \, dy = 0$. Rearranging the terms,we get: $(1 - e^{x}) \sec^{2} y \, dy = -e^{x} 	an y \, dx$. Separating the variables,we get: $\frac{\sec^{2} y}{	an y} \, dy = \frac{-e^{x}}{1 - e^{x}} \, dx$. Integrating both sides: $\int \frac{\sec^{2} y}{	an y} \, dy = \int \frac{-e^{x}}{1 - e^{x}} \, dx$. For the left side,let $	an y = u$,then $\sec^{2} y \, dy = du$. Thus,$\int \frac{du}{u} = \ln|u| = \ln|	an y|$. For the right side,let $1 - e^{x} = t$,then $-e^{x} \, dx = dt$. Thus,$\int \frac{dt}{t} = \ln|t| = \ln|1 - e^{x}|$. Combining these,we get: $\ln|	an y| = \ln|1 - e^{x}| + \ln|C|$. Using logarithmic properties: $\ln|	an y| = \ln|C(1 - e^{x})|$. Taking the exponential of both sides,we obtain the general solution: $	an y = C(1 - e^{x})$.

Find the general solution of the differential equation: $e^{x} \tan y \, dx + (1 - e^{x}) \sec^{2} y \, dy = 0$.

Similar Questions

For the differential equation $x y \frac{dy}{dx} = (x+2)(y+2)$,find the solution curve passing through the point $(1, -1)$.

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

Find the particular solution of the differential equation $\log \left(\frac{d y}{d x}\right)=3 x+4 y$ given that $y=0$ when $x=0$.

The general solution of $\frac{dy}{dx} = \cos^2(x-y-1)$ is given by $x=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module