### Core Logic Let's rearrange the given differential equation by expressing fracdydx$\frac{dy}{dx}$: x^5 fracdydx = 7x^4 cot y - e^x csc y$x^5 \frac{dy}{dx} = 7x^4 \cot y - e^x \csc y$ Dividing both sides by x^5$x^5$: fracdydx = frac7x cot y - frace^xx^5 csc y$\frac{dy}{dx} = \frac{7}{x} \cot y - \frac{e^x}{x^5} \csc y$ Multiply the entire equation by sin y$\sin y$ to clear denominators: sin y fracdydx - frac7x cos y = -frace^xx^5$\sin y \frac{dy}{dx} - \frac{7}{x} \cos y = -\frac{e^x}{x^5}$ This can be transformed into a linear form by substituting t = -cos y$t = -\cos y$. Then fracdtdx = sin y fracdydx$\frac{dt}{dx} = \sin y \frac{dy}{dx}$. ### Step 1: Solving the Linear ODE Substituting t$t$ leads to: fracdtdx + frac7x t = -frace^xx^5$\frac{dt}{dx} + \frac{7}{x} t = -\frac{e^x}{x^5}$ This is a standard linear first-order ODE with P(x) = frac7x$P(x) = \frac{7}{x}$. The Integrating Factor (I.F.) is: textI.F. = e^int frac7x dx = e^7 ln x = x^7$\text{I.F.} = e^{\int \frac{7}{x} dx} = e^{7 \ln x} = x^7$ The general solution is: t cdot x^7 = int left(-frace^xx^5right) cdot x^7 dx = -int x^2 e^x dx$t \cdot x^7 = \int \left(-\frac{e^x}{x^5}\right) \cdot x^7 dx = -\int x^2 e^x dx$ ### Step 2: Evaluating the Integration and Constant Using integration by parts for int x^2 e^x dx$\int x^2 e^x dx$: int x^2 e^x dx = x^2 e^x - 2xe^x + 2e^x$\int x^2 e^x dx = x^2 e^x - 2xe^x + 2e^x$ Substituting this back: -cos y cdot x^7 = -e^x(x^2 - 2x + 2) + C$-\cos y \cdot x^7 = -e^x(x^2 - 2x + 2) + C$ cos y cdot x^7 = e^x(x^2 - 2x + 2) - C$\cos y \cdot x^7 = e^x(x^2 - 2x + 2) - C$ Since the curve passes through left(1, fracpi2right)$\left(1, \frac{\pi}{2}\right)$: cosleft(fracpi2right) cdot (1)^7 = e^1(1^2 - 2(1) + 2) - C implies 0 = e(1) - C implies C = e$\cos\left(\frac{\pi}{2}\right) \cdot (1)^7 = e^1(1^2 - 2(1) + 2) - C \implies 0 = e(1) - C \implies C = e$ ### Step 3: Calculating cos y at x = 2 Now substitute x = 2$x = 2$ and C = e$C = e$ into our equation block: cos y cdot (2^7) = e^2(2^2 - 2(2) + 2) - e$\cos y \cdot (2^7) = e^2(2^2 - 2(2) + 2) - e$ cos y cdot 128 = e^2(4 - 4 + 2) - e = 2e^2 - e$\cos y \cdot 128 = e^2(4 - 4 + 2) - e = 2e^2 - e$ cos y = frac2e^2 - e128$\cos y = \frac{2e^2 - e}{128}$ ### Pattern Recognition When trigonometric terms are mixed inside an ODE containing derivative blocks like fracdxdy$\frac{dx}{dy}$ or fracdydx$\frac{dy}{dx}$, check if clearing denominators using sin y$\sin y$ or cos y$\cos y$ reveals a standard substitution path for a Linear ODE. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations