The particular solution of the differential e | vedclass

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

Question

(B) Given the differential equation: $\frac{dy}{dx} = y 	an x$. Separating the variables,we get: $\frac{dy}{y} = 	an x \, dx$. Integrating both side

Vedclass · Accepted Answer

$y = \sec x$

Vedclass · Answer

(B) Given the differential equation: $\frac{dy}{dx} = y 	an x$. Separating the variables,we get: $\frac{dy}{y} = 	an x \, dx$. Integrating both sides: $\int \frac{dy}{y} = \int 	an x \, dx$. This gives: $\ln |y| = \ln |\sec x| + C$. Exponentiating both sides: $|y| = e^{\ln |\sec x| + C} = e^C \cdot |\sec x|$. Let $e^C = k$,so $y = k \sec x$. Using the initial condition $y(0) = 1$: $1 = k \sec(0) \implies 1 = k(1) \implies k = 1$. Substituting $k = 1$ back into the equation,we get the particular solution: $y = \sec x$.

The particular solution of the differential equation $\frac{dy}{dx} = y \tan x$ with the initial condition $y(0) = 1$ is:

Similar Questions

The particular solution of the differential equation $\frac{dy}{dx} = \sec y$ with the initial condition $y(0) = 0$ is:

The solution of the differential equation $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$ is

Find a particular solution of the differential equation $(x-y)(dx+dy)=dx-dy$ given that $y=-1$ when $x=0$. (Hint: put $x-y=t$)

If $\frac{a}{a_1}=\frac{b}{b_1}$,then the substitution to be used to solve the differential equation $\frac{dy}{dx}=\frac{ax+by+c}{a_1x+b_1y+c_1}$ by using separation of variables is

The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module