Let the solution curve x=x(y), 0

Question

(A) The given differential equation is $(\ln(\cos y))^2 \cos y dx = (1+3x \ln(\cos y)) \sin y dy$.Rearranging the terms,we get $\frac{dx}{dy} = \frac{

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The given differential equation is $(\ln(\cos y))^2 \cos y dx = (1+3x \ln(\cos y)) \sin y dy$.Rearranging the terms,we get $\frac{dx}{dy} = \frac{(1+3x \ln(\cos y)) \sin y}{(\ln(\cos y))^2 \cos y} = 	an y \left( \frac{3x}{\ln(\cos y)} + \frac{1}{(\ln(\cos y))^2} ight)$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{3 	an y}{\ln(\cos y)}$ and $Q(y) = \frac{	an y}{(\ln(\cos y))^2}$.The integrating factor $IF = e^{\int P(y) dy} = e^{\int -\frac{3 	an y}{\ln(\cos y)} dy}$.Let $t = \ln(\cos y)$,then $dt = -	an y dy$. Thus,$IF = e^{\int \frac{3}{t} dt} = e^{3 \ln t} = t^3 = (\ln(\cos y))^3$.The solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.$x (\ln(\cos y))^3 = \int \frac{	an y}{(\ln(\cos y))^2} \cdot (\ln(\cos y))^3 dy + C = \int 	an y \ln(\cos y) dy + C$.Using $t = \ln(\cos y)$,$\int t (-dt) = -\frac{t^2}{2} + C$.So,$x (\ln(\cos y))^3 = -\frac{(\ln(\cos y))^2}{2} + C$.Given $x(\frac{\pi}{3}) = \frac{1}{2 \ln 2}$,we know $\cos(\frac{\pi}{3}) = \frac{1}{2}$,so $\ln(\cos(\frac{\pi}{3})) = \ln(\frac{1}{2}) = -\ln 2$.Substituting: $\frac{1}{2 \ln 2} (-\ln 2)^3 = -\frac{(-\ln 2)^2}{2} + C \implies -\frac{(\ln 2)^2}{2} = -\frac{(\ln 2)^2}{2} + C \implies C = 0$.Thus,$x = -\frac{1}{2 \ln(\cos y)}$.For $y = \frac{\pi}{6}$,$\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,so $x = -\frac{1}{2 \ln(\frac{\sqrt{3}}{2})} = -\frac{1}{2 (\frac{1}{2} \ln 3 - \ln 2)} = \frac{1}{\ln 4 - \ln 3} = \frac{1}{\ln(\frac{4}{3})}$.Comparing with $\frac{1}{\ln m - \ln n}$,we get $m=4, n=3$. Since $4$ and $3$ are co-prime,$mn = 12$.

Similar Questions

An integrating factor of the differential equation $(x^2+1) \frac{dy}{dx} + xy = x^3$ 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

Let $f : (0, \infty) \to (2, 20)$ be a twice differentiable function such that $\lim_{x \to \infty} (f(x) + f'(x) + f''(x)) = \lim_{x \to \infty} g(x)$,where $\lim_{x \to \infty} g(x)$ exists and is equal to $5$. Then $\lim_{x \to \infty} (f(x) - g(x))$ is equal to:

Let $y = y(x)$ be the solution curve of the differential equation $x(x^2 + e^x) dy + (e^x(x-2)y - x^3) dx = 0, x > 0$,passing through the point $(1, 0)$. Then $y(2)$ is equal to:

The integrating factor of the differential equation $(1-x^2) \frac{dy}{dx} + xy = kx$ for $(-1 < x < 1)$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module