Let y = y(x) be the solution of the different | vedclass

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

Question

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 	an x$ and $Q = 2x + x^2 	an x

Vedclass · Accepted Answer

$y'\left( \frac{\pi}{4} \right) - y'\left( -\frac{\pi}{4} \right) = \pi - \sqrt{2}$

Vedclass · Answer

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 	an x$ and $Q = 2x + x^2 	an x$.The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int 	an x dx} = e^{\ln |\sec x|} = \sec x$.Multiplying both sides by the $I$.$F$.,we get $y \sec x = \int (2x + x^2 	an x) \sec x dx$.$y \sec x = \int 2x \sec x dx + \int x^2 \sec x 	an x dx$.Using integration by parts for the first integral: $\int 2x \sec x dx$ is complex,but notice that $\frac{d}{dx}(x^2 \sec x) = 2x \sec x + x^2 \sec x 	an x$.Thus,$y \sec x = x^2 \sec x + C$.Dividing by $\sec x$,we get $y = x^2 + C \cos x$.Given $y(0) = 1$,we have $1 = 0^2 + C \cos(0) \Rightarrow C = 1$.So,$y = x^2 + \cos x$.Now,$y' = 2x - \sin x$.$y'\left( \frac{\pi}{4} ight) = 2\left( \frac{\pi}{4} ight) - \sin\left( \frac{\pi}{4} ight) = \frac{\pi}{2} - \frac{1}{\sqrt{2}}$.$y'\left( -\frac{\pi}{4} ight) = 2\left( -\frac{\pi}{4} ight) - \sin\left( -\frac{\pi}{4} ight) = -\frac{\pi}{2} + \frac{1}{\sqrt{2}}$.$y'\left( \frac{\pi}{4} ight) - y'\left( -\frac{\pi}{4} ight) = \left( \frac{\pi}{2} - \frac{1}{\sqrt{2}} ight) - \left( -\frac{\pi}{2} + \frac{1}{\sqrt{2}} ight) = \pi - \frac{2}{\sqrt{2}} = \pi - \sqrt{2}$.Thus,option $B$ is correct.

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x$,$x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$,such that $y(0) = 1$. Then

Similar Questions

Let $f$ be a differentiable function with $\lim_{x \rightarrow \infty} f(x) = 0$. If $y^{\prime} + y f^{\prime}(x) - f(x) f^{\prime}(x) = 0$ and $\lim_{x \rightarrow \infty} y(x) = 0$,then (where $y^{\prime} = \frac{dy}{dx}$):

Find the general solution of the differential equation $y dx - (x + 2y^2) dy = 0$.

If $y=y(x)$ is the solution curve of the differential equation $x^{2} dy + (y - \frac{1}{x}) dx = 0$ for $x > 0$ and $y(1) = 1$,then $y(\frac{1}{2})$ is equal to:

Let $y=y(x)$ satisfy the equation $\frac{dy}{dx}-|A|=0$,for all $x>0$,where $A=\begin{bmatrix} y & \sin x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x} \end{bmatrix}$. If $y(\pi)=\pi+2$,then the value of $y\left(\frac{\pi}{2}\right)$ is:

If the solution curve of the differential equation $(\tan^{-1} y - x) dy = (1 + y^2) dx$ passes through the point $(1, 0)$,then the abscissa of the point on the curve whose ordinate is $\tan(1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module