Let the solution curve $x=x(y)$, $0<y<\frac{\pi}{2}$, of the differential equation $(\log_{e}(\cos y))^{2} \cos y dx-(1+3x\log_{e}(\cos y))\sin y dy=0$ satisfy $x\left(\frac{\pi}{3}\right)=\frac{1}{2\log_{e} 2}$. If $x\left(\frac{\pi}{6}\right)=\frac{1}{\log_{e} m-\log_{e} n}$, where m and n are coprime, then mn is equal to ______.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Differential Equations

Let the solution curve

x = x (y)

0 < y < \frac{π}{2}

, of the differential equation

(lo g_{e} (cos y))^{2} cos y d x - (1 + 3 x lo g_{e} (cos y)) sin y d y = 0

satisfy

x (\frac{π}{3}) = \frac{1}{2 l o g _{e} 2}

. If

x (\frac{π}{6}) = \frac{1}{l o g _{e} m - l o g _{e} n}

, where m and n are coprime, then mn is equal to ______.

Answer: 12

Rewrite the differential equation as:

\frac{d x}{d y} - \frac{3 sin y}{cos y ( lo g _{e} cos y )} x = \frac{sin y}{( lo g _{e} cos y ) ^{2} \cdot cos y}

This is a linear first-order ODE. The integrating factor is:

I.F. = e^{\int \frac{- 3 s i n y}{c o s y ( l o g _{e} c o s y )} d y} = (ln cos y)^{3}

Multiplying and integrating:

x (ln cos y)^{3} = \int \frac{sin y \cdot ( ln cos y )}{cos y} d y = \frac{- ( ln cos y ) ^{2}}{2} + c

Using the initial condition

x (\frac{π}{3}) = \frac{1}{2 l o g _{e} 2}

\frac{1}{2 lo g _{e} 2} \cdot (ln \frac{1}{2})^{3} = \frac{- ( ln \frac{1}{2} ) ^{2}}{2} + c

This gives

c = 0

, so

x = \frac{- 1}{2 l n ( c o s y )}

.

At

y = \frac{π}{6}

x (\frac{π}{6}) = \frac{- 1}{2 l n ( \frac{3}{2} )} = \frac{1}{l n 4 - l n 3}

Thus

m = 4

n = 3

, and

mn = 12

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Integer answer