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 differential equation is $\frac{dy}{dx} - 2y 	an 2x = e^x \sec 2x$. This is a linear differential equation of the form $\frac{dy}{dx} +

Vedclass · Accepted Answer

$y \cos 2x = e^x + C$

Vedclass · Answer

(B) The given differential equation is $\frac{dy}{dx} - 2y 	an 2x = e^x \sec 2x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = -2 	an 2x$ and $Q = e^x \sec 2x$. The integrating factor $(IF)$ is given by $IF = e^{\int P dx} = e^{\int -2 	an 2x dx}$. $IF = e^{-\ln(\sec 2x)} = e^{\ln(\cos 2x)} = \cos 2x$. The general solution is $y \cdot (IF) = \int Q \cdot (IF) dx + C$. Substituting the values, we get $y \cos 2x = \int (e^x \sec 2x) \cdot \cos 2x dx + C$. Since $\sec 2x \cdot \cos 2x = 1$, we have $y \cos 2x = \int e^x dx + C$. Therefore, the solution is $y \cos 2x = e^x + C$, where $C$ is the constant of integration.

The solution of the differential equation $\frac{dy}{dx} - 2y \tan 2x = e^x \sec 2x$ is

Similar Questions

$\mathop {\lim}\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}} = $

When an orange solution containing $Cr_2O_7^{2-}$ ions is treated with an alkali,a yellow solution is formed,and when $H^{+}$ ions are added to the yellow solution,an orange solution is obtained. Explain why this happens.

There are $9$ balls to be placed in $9$ boxes. If $5$ of the balls cannot fit into $3$ specific small boxes,find the number of ways of arranging one ball in each of the boxes.

The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperature

If the lines $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$ are both tangents to a circle,then its radius is

Vedclass Test Series

Exam Paper Generator

Online Exam Module