Question

Solve the differential equation: $\cos x\frac{dy}{dx}+y\sin x={\sec }^{2}x$

• A. $y\sec x=-\tan x-\frac{1}{3}{\tan }^{3}x+c.$
• B. $y\sec x=\int (1+{\tan }^{2}x){\sec }^{2}xdx$
• C. $y\sec x=\tan x+\frac{1}{3}{\tan }^{3}x+c.$
• D. $y\sec x=-\int (1+{\tan }^{2}x){\sec }^{2}xdx$

Answer 1

Answer: (B), (C)

$y\sec x=\tan x+\frac{1}{3}{\tan }^{3}x+c.$
$y\sec x=\int (1+{\tan }^{2}x){\sec }^{2}xdx$

$\cos x\frac{dy}{dx}+y\sin x={\sec }^{2}x\Rightarrow \frac{dy}{dx}+y\tan x={\sec }^{3}x$ ...(1)
Here $P=\tan x\Rightarrow \int Pdx=\int \tan xdx=\log \sec x$
$\therefore I.F.=e^{\sec x}=\sec x$
Multiplying (1) by $I.F.$ we get
$\sec x\frac{dy}{dx}+y\tan x\sec x={\sec }^{4}x$
Integrating both sides we get
$y\sec x=\int {\sec }^{4}xdx+c=\int (1+{\tan }^{2}x){\sec }^{2}xdx=\tan x+\frac{1}{3}{\tan }^{3}x+c$