Answer to Question #154230 in Differential Equations for Jishnu

Question

Show that y=x+tanx satisfies the differential equation cos²x d²y /dx² -2y+2x=0

Answer 1

Given, $y=x+\tan x ...(i)$

On differentiating both sides w.r.t $x$ ,

$\frac{dy}{dx}=1+\sec ^2x$

Again, differentiating both sides w.r.t $x$ ,

$\frac{d^2y}{dx^2}=0+2\sec x \frac{d (\sec x)}{dx}=2\sec x \sec x\tan x \\\Rightarrow \frac{d^2y}{dx^2}=2\sec^2 x \tan x ...(ii)$

Now, differential equation is: $\cos ^2x \frac{d^2y}{dx^2}-2y+2x=0$

Put values from equation (i) and (ii) into LHS of this, we get:

$=\cos ^2x (2\sec^2 x \tan x)-2(x+\tan x)+2x$

$=\cos ^2x (2 \times\frac{1}{\cos ^2x} \tan x)-2x-2\tan x+2x \ [\because \ \sec x=\frac{1}{\cos x}] \\ =2 \tan x-2x-2\tan x+2x \\ =0+0 \\ =0 \\ =RHS$

Hence, proved.