A particular solution of 3 e^x tan y , dx + ( | vedclass

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

Question

(D) Given the differential equation: $3 e^x 	an y \, dx + (1 - e^x) \sec^2 y \, dy = 0$. Rearranging the terms to separate variables: $\frac{\sec^2 y

Vedclass · Accepted Answer

$	an y = \left(\frac{1 - e^x}{1 - e}ight)^3$

Vedclass · Answer

(D) Given the differential equation: $3 e^x 	an y \, dx + (1 - e^x) \sec^2 y \, dy = 0$. Rearranging the terms to separate variables: $\frac{\sec^2 y}{	an y} \, dy = -\frac{3 e^x}{1 - e^x} \, dx$. Integrating both sides: $\int \frac{\sec^2 y}{	an y} \, dy = \int \frac{3 e^x}{e^x - 1} \, dx$. Let $u = 	an y$,then $du = \sec^2 y \, dy$. The left side becomes $\ln|	an y|$. For the right side,let $v = e^x - 1$,then $dv = e^x \, dx$. The right side becomes $3 \ln|e^x - 1| + C$. So,$\ln|	an y| = 3 \ln|e^x - 1| + C = \ln|e^x - 1|^3 + C$. This implies $	an y = K(e^x - 1)^3$. Given $y(1) = \frac{\pi}{4}$,we have $	an(\frac{\pi}{4}) = K(e^1 - 1)^3$,so $1 = K(e - 1)^3$,which gives $K = \frac{1}{(e - 1)^3}$. Substituting $K$ back: $	an y = \frac{(e^x - 1)^3}{(e - 1)^3} = \left(\frac{e^x - 1}{e - 1}ight)^3$. Since $(e^x - 1)^3 = -(1 - e^x)^3$ and $(e - 1)^3 = -(1 - e)^3$,we get $	an y = \left(\frac{1 - e^x}{1 - e}ight)^3$.

$A$ particular solution of $3 e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0$ with $y(1) = \frac{\pi}{4}$ is

Similar Questions

The equation of the curve passing through the point $(1,0)$ and whose slope is $\frac{y - 1}{x^2 + x}$ is

The solution of $\frac{dy}{dx} = x \log x$ is

The general solution of the differential equation $\frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}}$ is

The solution of the differential equation $x \frac{dy}{dx} - y = 3$ represents a family of

If $\frac{dy}{dx} + \frac{2^{x-y}(2^y - 1)}{2^x - 1} = 0$,$x, y > 0$,and $y(1) = 1$,then $y(2)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module