Question

Solve d 2y dx 2 − 2 tan x dy dx + 5y = 0

Answer 1

Answer:

solve: d^2/dx^2 - 2 tan x dy/dx + 5y = e^x sec x

Answer 2

Solve the given differential equation $\frac{d^2y}{dx^2}-2tanx \frac{dy}{dx}+5y=0$

Explanation:

• In order to solve a differential equation we need to eliminate all the derivative terms from the equation and simply obtain a function in 'x' and 'y'.
• Given is the differential equation, $\frac{d^2y}{dx^2}-2tanx \frac{dy}{dx}+5y=0$
• Multiplying both sides by $cosx$ in the given equation we get,     $->cosx\frac{d^2y}{dx^2}-2sinx \frac{dy}{dx}+5ycosx=0$    ---(i)
• Now let us consider another function 'u' such that ,
•                          $u=ycosx$      ---(a)
• Differentiating 'u' we get,  
•                  $->\frac{du}{dy}=cosx\frac{dy}{dx} -ysinx\\->\frac{d^2u}{dy^2}=cosx\frac{d^2y}{dx^2} -2\frac{dy}{dx} sinx-ycosx$  ---(ii)
• From (i) and (ii) we get,            $\frac{d^2u}{dx^2}+6u=0$
• To solve this we put $u=e^{kx}$ hence, we get
• $\frac{d^2(e^{kx})}{dx^2}+6e^{kx}=0\\ ->k^2e^{kx}+6e^{kx}=0\\->(k^2+6)e^{kx}=0\\->k=\pm i\sqrt{6} \\->u=c\ e^{i\sqrt{6} }+c'e^{-i\sqrt{6} }$    (here c and c' are arbitrary constant)  
• using Euler identity we can further simplify u as,
• $->u=c(cos(x\sqrt{6} )+isin(x\sqrt{6} ))+c'(cos(x\sqrt{6} )-isin(x\sqrt{6} ))\\->u=(c+c')cos(x\sqrt{6} )+(c-c')isin(x\sqrt{6} )\\->u=kcos(x\sqrt{6} )+k'isin(x\sqrt{6} )$      
• Here k and k' are arbitrary constants.                          
• From (a) we get the final solution of the differential equation as,   $y=\frac{1}{cosx}(kcos(x\sqrt{6} )+k'isin(x\sqrt{6} ))$