If y(x) satisfies the differential equation y | vedclass

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

Question

(A) The given differential equation is $\frac{dy}{dx} - y 	an x = 2x \sec x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py =

Vedclass · Accepted Answer

($A$, $D$)

Vedclass · Answer

(A) The given differential equation is $\frac{dy}{dx} - y 	an x = 2x \sec x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = -	an x$ and $Q = 2x \sec x$.The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{-\int 	an x dx} = e^{-\ln(\sec x)} = \cos x$.Multiplying both sides by the $I.F.$, we get $\frac{d}{dx}(y \cos x) = 2x \sec x \cdot \cos x = 2x$.Integrating both sides, $y \cos x = x^2 + C$.Given $y(0) = 0$, we have $0 = 0^2 + C$, so $C = 0$.Thus, $y = x^2 \sec x$.For option $(A)$: $y(\frac{\pi}{4}) = (\frac{\pi}{4})^2 \sec(\frac{\pi}{4}) = \frac{\pi^2}{16} \cdot \sqrt{2} = \frac{\pi^2}{8\sqrt{2}}$. This is true.For option $(D)$: $y'(x) = 2x \sec x + x^2 \sec x 	an x$.$y'(\frac{\pi}{3}) = 2(\frac{\pi}{3}) \sec(\frac{\pi}{3}) + (\frac{\pi}{3})^2 \sec(\frac{\pi}{3}) 	an(\frac{\pi}{3}) = 2(\frac{\pi}{3})(2) + \frac{\pi^2}{9}(2)(\sqrt{3}) = \frac{4\pi}{3} + \frac{2\pi^2\sqrt{3}}{9}$. This is true.Therefore, options $(A)$ and $(D)$ are correct.

Similar Questions

The solution of the differential equation $\frac{dy}{dx} + \frac{y}{x \log_{e} x} = \frac{1}{x}$ under the condition $y = 1$ when $x = e$ is

If $\frac{dy}{dx} + y \tan x = \sin 2x$ and $y(0) = 1$,then $y(\pi)$ is equal to

The curve satisfying the differential equation $y \, dx - (x + 3y^2) \, dy = 0$ and passing through the point $(1, 1)$ also passes through the point

The solution of $\frac{dy}{dx} + p(x)y = 0$ is

Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} + \left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y = \frac{x+3}{x+1}$,where $x > -1$,which passes through the point $(0,1)$. Then $y(1)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module