Let y = y(x)$y = y(x)$ be the solution of the differential equation sec xfracdydx - 2y = 2 + 3sin x, x in left(-fracpi2, fracpi2right)$\sec x\frac{dy}{dx} - 2y = 2 + 3\sin x, x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, y(0) = -frac74$y(0) = -\frac{7}{4}$. Then yleft(fracpi6right)$y\left(\frac{\pi}{6}\right)$ is equal to:

### Related Formula textStandard form LDE: fracdydx + P(x)y = Q(x)$$\text{Standard form LDE: } \frac{dy}{dx} + P(x)y = Q(x)$$ textIntegrating Factor (I.F.) = e^int P(x)dx$$\text{Integrating Factor (I.F.) } = e^{\int P(x)dx}$$ textSolution is y(I.F.) = int Q(x)(I.F.)dx + C$$\text{Solution is } y(I.F.) = \int Q(x)(I.F.)dx + C$$ ### Core Logic Multiply the entire differential equation by cos x$\cos x$ to normalize the fracdydx$\frac{dy}{dx}$ term: fracdydx - (2cos x)y = 2cos x + 3sin xcos x$$\frac{dy}{dx} - (2\cos x)y = 2\cos x + 3\sin x\cos x$$ This is a standard first-order linear differential equation with P(x) = -2cos x$P(x) = -2\cos x$. ### Step 1: Integrating Factor textI.F. = e^int -2cos x \, dx = e^-2sin x$$\text{I.F.} = e^{\int -2\cos x \, dx} = e^{-2\sin x}$$ ### Step 2: General Solution y cdot e^-2sin x = int e^-2sin x (2cos x + 3sin xcos x) \, dx$$y \cdot e^{-2\sin x} = \int e^{-2\sin x} (2\cos x + 3\sin x\cos x) \, dx$$ Let u = sin x implies du = cos x \, dx$u = \sin x \implies du = \cos x \, dx$. The integral becomes: int e^-2u (2 + 3u) \, du$$\int e^{-2u} (2 + 3u) \, du$$ Using integration by parts: = (2+3u)left(frace^-2u-2right) - int 3 left(frace^-2u-2right) \, du$$= (2+3u)\left(\frac{e^{-2u}}{-2}\right) - \int 3 \left(\frac{e^{-2u}}{-2}\right) \, du$$ = -frac2+3u2 e^-2u - frac34 e^-2u$$= -\frac{2+3u}{2} e^{-2u} - \frac{3}{4} e^{-2u}$$ = e^-2u left(-frac2+3u2 - frac34right) = e^-2u left(-frac4+6u+34right) = e^-2u left(-frac6u+74right)$$= e^{-2u} \left(-\frac{2+3u}{2} - \frac{3}{4}\right) = e^{-2u} \left(-\frac{4+6u+3}{4}\right) = e^{-2u} \left(-\frac{6u+7}{4}\right)$$ Re-substitute u = sin x$u = \sin x$: y e^-2sin x = e^-2sin x left(-frac32sin x - frac74right) + C$$y e^{-2\sin x} = e^{-2\sin x} \left(-\frac{3}{2}\sin x - \frac{7}{4}\right) + C$$ y = -frac32sin x - frac74 + C e^2sin x$$y = -\frac{3}{2}\sin x - \frac{7}{4} + C e^{2\sin x}$$ ### Step 3: Apply Boundary Conditions Given y(0) = -frac74$y(0) = -\frac{7}{4}$: -frac74 = -frac32(0) - frac74 + C e^0 implies C = 0$$-\frac{7}{4} = -\frac{3}{2}(0) - \frac{7}{4} + C e^0 \implies C = 0$$ Thus, y(x) = -frac32sin x - frac74$y(x) = -\frac{3}{2}\sin x - \frac{7}{4}$. ### Step 4: Find Final Value Calculate yleft(fracpi6right)$y\left(\frac{\pi}{6}\right)$: yleft(fracpi6right) = -frac32sinleft(fracpi6right) - frac74 = -frac32left(frac12right) - frac74$$y\left(\frac{\pi}{6}\right) = -\frac{3}{2}\sin\left(\frac{\pi}{6}\right) - \frac{7}{4} = -\frac{3}{2}\left(\frac{1}{2}\right) - \frac{7}{4}$$ = -frac34 - frac74 = -frac104 = -frac52$$= -\frac{3}{4} - \frac{7}{4} = -\frac{10}{4} = -\frac{5}{2}$$ ### Pattern Recognition When integrating e^ax(f(x))$e^{ax}(f(x))$, use DI method or standard Integration by Parts mapping. For int e^-2u(3u+2)$\int e^{-2u}(3u+2)$, always extract polynomial as u$u$, exponential as dv$dv$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations