If a curve y=y(x) passes through (1,2) and satisfies (7x^4 y-e^x y)dxdy=x^5, x1, then at x…

Question

If a curve $y=y(x)$ passes through $\left(1,\dfrac{\pi}{2}\right)$ and satisfies $(7x^4\cot y-e^x\csc y)\dfrac{dx}{dy}=x^5$, $x\ge1$, then at $x=2$, the value of $\cos y$ is:

Accepted Answer

$$x^5\frac{dy}{dx}-7x^4\cot y = -e^x\csc y \implies \frac{dy}{dx}-\frac7x\cot y = -\frac{e^x}{x^5}\csc y.$$

Multiplying by $\sin y$ and setting $t=-\cos y$:
$$\frac{dt}{dx}+\frac7xt = -\frac{e^x}{x^5}.$$

I.F. $=x^7$.
$$tx^7 = -\int e^xx^2\,dx = -e^x(x^2-2x+2)+c.$$

Using $x=1,y=\pi/2 \Rightarrow t=0$: $c=e$. So $-\cos y\cdot x^7 = -e^x(x^2-2x+2)+e$, i.e. $\cos y\cdot x^7 = e^x(x^2-2x+2)-e$.

At $x=2$: $\cos y = \dfrac{2e^2-e}{128}$.

Solutions · 1