# Differential equation with sec

With $(a)$ I got that $-y^2 dx = \sec^2x\ dy$, but it makes no sense. Hence, no Idea how to handle $(b)$.

• For $(i)$, since $0\le x\le\dfrac\pi2$, I think we can use the facts that $\cos x\le1$ and $\sin x +\cos x\ge1$. Divide them by $\cos x$ and combine. – Tunk-Fey Jun 14 '14 at 22:16

For $(a)$, using separation of variables, we obtain \begin{align} \frac{dy}{y^2}&=-\frac1{\sec^2x}\ dx\\ \int y^{-2}\ dy&=-\int \cos^2x\ dx\\ \int y^{-2}\ dy&=-\int\frac{1+\cos 2x}{2}\ dx. \end{align}
• For $(i)$, since $0\le x\le\dfrac\pi2$, I think we can use the facts that $\cos x\le1$ and $\sin x +\cos x\ge1$. Divide them by $\cos x$ and combine. – Tunk-Fey Jun 14 '14 at 22:19
$$\int \cos^2(x)dx = \int \frac{-1}{y^2}dy$$
I simply took what you already had and divided both sides by $-y^2$ and $\sec^2(x)$. Now it's just a matter of good old integration. :)