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

Find the general solution of the differential equation: $\frac{dy}{dx} = \sin^{-1} x$

The solution of the differential equation $\frac{dy}{dx} = (4x + y + 1)^2$, when $y(0) = 1$ is

If $y(x)$ is the solution of the differential equation $(x+2) \frac{dy}{dx} = x^2+4x-9, x \neq -2$ and $y(0) = 0$,then $y(-4)$ is equal to

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

If a curve passes through the point $(1,1)$ and at any point $(x, y)$ on the curve,the product of the slope of its tangent and $x$ coordinate of the point is equal to the $y$ coordinate of the point,then the curve also passes through the point

Vedclass Test Series

Exam Paper Generator

Online Exam Module