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(C) Given the differential equation: $\left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5$. Rearranging the equation: $\frac{d x}

Vedclass · Accepted Answer

$\frac{2 e^2-e}{128}$

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(C) Given the differential equation: $\left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5$. Rearranging the equation: $\frac{d x}{d y} = \frac{x^5}{7 x^4 \cot y - e^x \operatorname{cosec} y}$. This is equivalent to $\frac{d y}{d x} = \frac{7 x^4 \cot y - e^x \operatorname{cosec} y}{x^5} = \frac{7 \cot y}{x} - \frac{e^x \operatorname{cosec} y}{x^5}$. Multiplying by $\sin y$: $\sin y \frac{d y}{d x} - \frac{7 \cos y}{x} = -\frac{e^x}{x^5}$. Let $t = \cos y$,then $\frac{d t}{d x} = -\sin y \frac{d y}{d x}$. Substituting this into the equation: $-\frac{d t}{d x} - \frac{7 t}{x} = -\frac{e^x}{x^5}$,which simplifies to $\frac{d t}{d x} + \frac{7 t}{x} = \frac{e^x}{x^5}$. This is a linear differential equation with Integrating Factor $I.F. = e^{\int \frac{7}{x} dx} = e^{7 \ln x} = x^7$. The solution is $t \cdot x^7 = \int x^7 \cdot \frac{e^x}{x^5} dx = \int x^2 e^x dx$. Using integration by parts: $\int x^2 e^x dx = x^2 e^x - 2 \int x e^x dx = x^2 e^x - 2(x e^x - e^x) + C = e^x(x^2 - 2x + 2) + C$. So,$\cos y \cdot x^7 = e^x(x^2 - 2x + 2) + C$. At $x=1, y=\frac{\pi}{2}$,we have $\cos(\frac{\pi}{2}) \cdot 1^7 = e^1(1-2+2) + C \implies 0 = e + C \implies C = -e$. Thus,$\cos y = \frac{e^x(x^2 - 2x + 2) - e}{x^7}$. At $x=2$,$\cos y = \frac{e^2(4 - 4 + 2) - e}{2^7} = \frac{2e^2 - e}{128}$.

If a curve $y=y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation $\left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5, x \geq 1$,then at $x=2$,the value of $\cos y$ is:

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