If a curve y=y(x) passes through the point (1,( )/(2)) and satisfies the differential equa…

Question

If a curve $y=y(x)$ passes through the point $(1,\frac{\pi }{2})$ and satisfies the differential equation $(7{x}^{4}\cot y-{e}^{x}\csc y)\frac{dx}{dy}={x}^{5},x\ge 1$, then at $x=2$, the value of $\cos y$ is :

Accepted Answer

$$\begin{aligned}
& \left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5 \
& x^5 \frac{d y}{d x}-7 x^4 \cot y=-e^x \operatorname{cosec} y \
& \frac{d y}{d x}-\frac{7}{x} \cot y=-\frac{e^x}{x^5} \operatorname{cosec} y \
& \sin y \frac{d y}{d x}-\frac{7}{x} \cos y=-\frac{e^x}{x^5} \
& \text { Let }-\cos y=t \
& \sin y \frac{d y}{d x}=\frac{d t}{d x} \
& \therefore \frac{d t}{d x}+\frac{7}{x} t=-\frac{e^x}{x^5}
\end{aligned}$$

$$\begin{aligned}
& \therefore \text { I.F. }=e^{\int \frac{7}{x} d x}=x^7 \
& t \cdot x^7=\int \frac{-e^x}{x^5} \cdot x^7 d x \
& -\cos y \cdot x^7=-\int e^x x^2 d x \
& \cos y x^7=e^x\left(x^2-2 x+2\right)+c \
& \because \quad x=1 \text { then } y=\frac{\pi}{2} \Rightarrow c=-e \
& \therefore \quad \cos y \cdot x^7=e^x\left(x^2-2 x+2\right)-e \
& \text { When } x=2 \text { then } \cos y=\frac{2 e^2-e}{128}
\end{aligned}$$

Solutions · 1