Answer to Question #154087 in Differential Equations for Sakshi Naik Gaonkar

Question

Compute W(y1,y2) where y1 and y2 are solutions of the differential equation y"+y'+y=0

Answer 1

y''+y'+y=0 is a linear homogeneous differential equation with constant coefficients.

The solution is sought in the form of an exponent y = $e^{\lambda x}$ .

We write the characteristic equation assuming that

y' = $\lambda$$e^{\lambda x}$ ; y'' = $\lambda$²$e^{\lambda x}$

Hence

$\lambda$²$e^{\lambda x}$ + $\lambda$$e^{\lambda x}$ + $e^{\lambda x}$ = 0 or $e^{\lambda x}$ ( $\lambda$² + $\lambda$ + 1) = 0

$\lambda$² + $\lambda$ + 1 = 0 is the caracteristic equation.

The above equation is the standard quadratic equation:

D = b² - 4ac; a=b=c=1

D = 1² -4*1*1 = -3

$\lambda$_1,2 = $\frac{ - b ± \sqrt{\smash[b]{D}}}{2a}$

$\lambda$₁ = $\frac{ - 1 + i \sqrt{\smash[b]{3}}}{2}$

$\lambda$₂ = $\frac{ - 1 - i \sqrt{\smash[b]{3}}}{2}$

y = c₁exp( $\lambda$₁x) + c₂exp( $\lambda$₂x)

y₁ = c₁exp( $\frac{ - 1 + i \sqrt{\smash[b]{3}}}{2}$x)

y₂ = c₂exp( $\frac{ - 1 - i \sqrt{\smash[b]{3}}}{2}$x)

If y = u(x)+iv(x) is a solution to the equation then y₁=u(x) and y₂=iv(x) (here i = $\sqrt-1$ ) are solutions too.

Hence:

we can write

y₁=c₁$e^\frac{-x}{2}$sin($\frac{\sqrt3}{2}$x);

y₂=c₂$e^\frac{-x}{2}$ cos($\frac{\sqrt3}{2}$x)

W = $\begin{vmatrix} y_1 & y_2 \\ y'_1 & y'_2 \end{vmatrix}$

y'₁ = c₁$\lbrace$ -$\frac{1}{2}$$e^\frac{-x}{2}$sin($\frac{\sqrt{\smash[b]{3}}}{2}$x) +$\frac{\sqrt{\smash[b]{3}}}{2}$$e^\frac{-x}{2}$cos($\frac{\sqrt{\smash[b]{3}}}{2}$x)$\rbrace$

y'₂ = c₂$\lbrace$ -$\frac{1}{2}$$e^\frac{-x}{2}$cos($\frac{\sqrt{\smash[b]{3}}}{2}$x) -$\frac{\sqrt{\smash[b]{3}}}{2}$$e^\frac{-x}{2}$sin($\frac{\sqrt{\smash[b]{3}}}{2}$x)$\rbrace$

W = $\begin{vmatrix} c_1e^\frac{-x}{2}sin(\frac{\sqrt3}{2}x) & c_2e^\frac{-x}{2}cos(\frac{\sqrt3}{2}x) \\ \frac{c_1e^\frac{-x}{2}}{2}\lbrace-sin(\frac{\sqrt3}{2}x)+\sqrt3cos(\frac{\sqrt3}{2}x)\rbrace & \frac{-c_2e^\frac{-x}{2}}{2}\lbrace cos(\frac{\sqrt3}{2}x)+\sqrt3sin(\frac{\sqrt3}{2}x)\rbrace \end{vmatrix}$ =

= $c_1e^\frac{-x}{2}sin(\frac{\sqrt3}{2}x)(\frac{-c_2e^\frac{-x}{2}}{2})\lbrace cos(\frac{\sqrt3}{2}x)+\sqrt3sin(\frac{\sqrt3}{2}x)\rbrace-$

$- c_2e^\frac{-x}{2}cos(\frac{\sqrt3}{2}x)(\frac{c_1e^\frac{-x}{2}}{2})\lbrace-sin(\frac{\sqrt3}{2}x)+\sqrt3cos(\frac{\sqrt3}{2}x)\rbrace =$

$= - \frac{c_1c_2}{2}e^{-x}\lbrace \bcancel{cos(\frac{\sqrt3}{2}x)sin(\frac{\sqrt3}{2}x)} - \bcancel{cos(\frac{\sqrt3}{2}x)sin(\frac{\sqrt3}{2}x)} + \\ +\sqrt3sin^2(\frac{\sqrt3}{2}x)+\sqrt3cos^2(\frac{\sqrt3}{2}x) \rbrace=$

$=- \frac{\sqrt3}{2}c_1c_2e^{-x}$

$if\\ c_1=c_2=1 \\W = =- \frac{\sqrt3}{2}e^{-x}$