If for the solution curve y=f(x) of the diffe | vedclass

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

Question

(A) The given differential equation is $\frac{dy}{dx}+(	an x)y=\frac{2+\sec x}{(1+2\sec x)^2}$.This is a linear differential equation of the form $\f

Vedclass · Accepted Answer

$\frac{4-\sqrt{2}}{14}$

Vedclass · Answer

(A) The given differential equation is $\frac{dy}{dx}+(	an x)y=\frac{2+\sec x}{(1+2\sec x)^2}$.This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=	an x$ and $Q=\frac{2+\sec x}{(1+2\sec x)^2}$.The integrating factor $IF = e^{\int 	an x dx} = e^{\ln|\sec x|} = \sec x$.The general solution is $y \cdot \sec x = \int Q \cdot IF dx + C$.$y \sec x = \int \frac{2+\sec x}{(1+2\sec x)^2} \sec x dx = \int \frac{2\cos x+1}{(\cos x+2)^2} dx$.Using the substitution $\cos x = \frac{1-t^2}{1+t^2}$ where $t = 	an(x/2)$,we get $dx = \frac{2dt}{1+t^2}$.$y \sec x = \int \frac{2(\frac{1-t^2}{1+t^2})+1}{(\frac{1-t^2}{1+t^2}+2)^2} \frac{2dt}{1+t^2} = \int \frac{3-t^2}{(t^2+3)^2} 2dt$.Let $u = t + \frac{3}{t}$,then $du = (1 - \frac{3}{t^2}) dt = \frac{t^2-3}{t^2} dt$. This integral simplifies to $y \sec x = \frac{2}{t+3/t} + C = \frac{2t}{t^2+3} + C$.Given $f(\pi/3) = \sqrt{3}/10$,at $x=\pi/3$,$t = 	an(\pi/6) = 1/\sqrt{3}$.$(\sqrt{3}/10) \cdot 2 = \frac{2(1/\sqrt{3})}{1/3+3} + C \implies \sqrt{3}/5 = \frac{2/\sqrt{3}}{10/3} + C = \frac{\sqrt{3}}{5} + C \implies C=0$.Thus $y \sec x = \frac{2t}{t^2+3}$. At $x=\pi/4$,$t = 	an(\pi/8) = \sqrt{2}-1$.$y \cdot \sqrt{2} = \frac{2(\sqrt{2}-1)}{(\sqrt{2}-1)^2+3} = \frac{2(\sqrt{2}-1)}{2-2\sqrt{2}+1+3} = \frac{2(\sqrt{2}-1)}{6-2\sqrt{2}} = \frac{\sqrt{2}-1}{3-\sqrt{2}}$.$y = \frac{\sqrt{2}-1}{\sqrt{2}(3-\sqrt{2})} = \frac{\sqrt{2}-1}{3\sqrt{2}-2} \cdot \frac{3\sqrt{2}+2}{3\sqrt{2}+2} = \frac{6+\sqrt{2}-2}{18-4} = \frac{4+\sqrt{2}}{14}$.Wait,re-evaluating the integral $\int \frac{3-t^2}{(t^2+3)^2} 2dt$: $y \sec x = \frac{2t}{t^2+3} + C$ is correct. With $C=0$,$y = \cos x \cdot \frac{2t}{t^2+3} = \frac{1-t^2}{1+t^2} \cdot \frac{2t}{t^2+3}$.At $t=\sqrt{2}-1$,$t^2 = 3-2\sqrt{2}$,$y = \frac{1-(3-2\sqrt{2})}{1+3-2\sqrt{2}} \cdot \frac{2(\sqrt{2}-1)}{3-2\sqrt{2}+3} = \frac{2\sqrt{2}-2}{4-2\sqrt{2}} \cdot \frac{2\sqrt{2}-2}{6-2\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}-1}{3-\sqrt{2}} = \frac{4-\sqrt{2}}{14}$.

If for the solution curve $y=f(x)$ of the differential equation $\frac{dy}{dx}+(\tan x)y=\frac{2+\sec x}{(1+2\sec x)^2}$,$x \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$,$f\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{10}$,then $f\left(\frac{\pi}{4}\right)$ is equal to:

Similar Questions

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

Find the solution of the differential equation $(e^{y-x}) dy = (e^x - e^y) dx$.

Let the solution curve $y = y(x)$ of the differential equation $\frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}} y = 2x \exp \left( \frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}} \right)$ pass through the origin. Then $y(1)$ is equal to:

Find the general solution of the differential equation: $x \frac{dy}{dx} + y - x + xy \cot x = 0$ $(x \neq 0)$.

The particular solution of the differential equation $\sin^{2} y \frac{dx}{dy} + x = \cot y$ when $x = 0$ and $y = \frac{3\pi}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module