The solution of the differential equation fra | vedclass

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

Question

(B) The given linear differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = -	an x$ and $Q = e^x \sec x$. First,we find the integra

Vedclass · Accepted Answer

$y \cos x = e^x + c$

Vedclass · Answer

(B) The given linear differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = -	an x$ and $Q = e^x \sec x$. First,we find the integrating factor $(IF)$: $IF = e^{\int P dx} = e^{\int -	an x dx} = e^{\ln(\cos x)} = \cos x$. The general solution is given by $y \cdot (IF) = \int Q \cdot (IF) dx + c$. Substituting the values: $y \cos x = \int (e^x \sec x) \cdot \cos x dx + c$. Since $\sec x \cdot \cos x = 1$,we have: $y \cos x = \int e^x dx + c$. $y \cos x = e^x + c$.

The solution of the differential equation $\frac{dy}{dx} - y \tan x = e^x \sec x$ is

Similar Questions

The solution of the differential equation $2 \frac{dy}{dx} - \frac{y}{x} = \frac{y^2}{x^2}$,given that $y = 2$ when $x = 1$,is

The solution of the differential equation $\frac{dx}{dy} = \frac{x}{1 + x e^y \cos(x^2)}$ is (where $c$ is the constant of integration):

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

If $y = y(x)$ is the solution of the differential equation $\frac{dy}{dx} + (\tan x)y = \sin x, 0 \leq x \leq \frac{\pi}{3},$ with $y(0) = 0,$ then $y\left(\frac{\pi}{4}\right)$ is equal to:

The integrating factor of the differential equation $(1 + t^2) + (x - e^{\tan^{-1} t}) \frac{dt}{dx} = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module