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: $(e^{x}+e^{-x}) dy = (e^{x}-e^{-x}) dx$. Separating the variables,we get: $dy = \frac{e^{x}-e^{-x}}{e^{x}+e^{-

Vedclass · Accepted Answer

$y = \log(e^{x} + e^{-x}) + C$

Vedclass · Answer

(A) The given differential equation is: $(e^{x}+e^{-x}) dy = (e^{x}-e^{-x}) dx$. Separating the variables,we get: $dy = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx$. Integrating both sides: $\int dy = \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx + C$. Let $t = e^{x}+e^{-x}$. Then $dt = (e^{x}-e^{-x}) dx$. Substituting these into the integral: $y = \int \frac{1}{t} dt + C$. Integrating,we get: $y = \log|t| + C$. Substituting back $t = e^{x}+e^{-x}$,the general solution is: $y = \log(e^{x}+e^{-x}) + C$.

Find the general solution of the differential equation: $(e^{x}+e^{-x}) dy - (e^{x}-e^{-x}) dx = 0$.

Similar Questions

The particular solution of the differential equation $(y + x \cdot \frac{dy}{dx}) \cdot \sin(xy) = \cos x$ at $x = 0$ is

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

The general solution of $\frac{dy}{dx} = \frac{x^3(y^4+1)}{\left[2y^{-2/3} + 3\left(\frac{x}{y^{1/3}}\right)^2\right]^{3/2}}$ is

$A$ curve having the condition that the slope of the tangent at any point $(x, y)$ is two times the slope of the straight line joining the same point to the origin $(0, 0)$ is a/an:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module