Question

Find the solution of the initial value problem y Superscript prime prime Baseline plus 4 y Superscript prime Baseline plus 5 y equals 0, y left-parenthesis StartFraction pi Over 2 EndFraction right-parenthesis equals 0 and y Superscript prime Baseline ⁢ left-parenthesis StartFraction pi Over 2 EndFraction right-parenthesis equals 5.

Answer 1

Answer:

$y(t) = - 5 {e}^{2\pi - 2t} \cos(t)$

Step-by-step explanation:

The given initial value problem is

$y''+4y'+5=0$

$y( \frac{ \pi}{2} ) = 0$

$y'( \frac{ \pi}{2} ) = 5$

The corresponding characteristic equation is

${m}^{2} + 4m + 5 = 0$

$m = - 2 \pm \: i$

The general solution becomes:

$y(t) = A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \sin(t)$

We differentiate to get:

$y'(t) = - A {e}^{ - 2t} \sin(t) - 2 A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \cos(t) - 2B {e}^{ - 2t} \sin(t)$

We apply the initial conditions to get;

$y( \frac{\pi}{2} ) = A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$

$A {e}^{ - 2\pi} (0 ) + B {e}^{ - 2\pi} ( 1) = 0$

$B = 0$

Also;

$y'( \frac{\pi}{2} ) = - A {e}^{ - 2\pi} \sin( \frac{\pi}{2} ) - 2 A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) - 2B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$

$- A {e}^{ - 2\pi} ( 1 ) - 2 A {e}^{ - 2\pi} ( 0 ) + B {e}^{ - 2\pi}( 0) - 2B {e}^{ - 2\pi} ( 1 ) = 5$

$- A {e}^{ - 2\pi} - 2B {e}^{ - 2\pi} = 5$

But B=0

$- A {e}^{ - 2\pi} = 5$

$A = 5{e}^{2\pi}$

Therefore the particular solution is

$y(t) = - 5 {e}^{2\pi} {e}^{ - 2t} \cos(t) + 0 \times {e}^{ - 2t} \sin(t)$

$y(t) = - 5 {e}^{2\pi - 2t} \cos(t)$