Answer to Question #122584 in Differential Equations for jse

Question

1. Find the general solution of the given higher-order differential equation.
16 d^4 y/ dx^4 + 24 d^2 y/ dx^2+ 9y = 0

y(x) =_____

2. Find the general solution of the given second-order differential equation.
y'' − 10y' + 26y = 0

y(x) = ______

Answer 1

1) Given differential equation is $16 \frac{d^4 y}{dx^4} + 24 \frac{d^2 y}{dx^2}+ 9y = 0$ .

$\implies (16D^4+24D^2+9)y = 0$ where $D = \frac{d}{dx}$ .

$\implies (4D^2+3)^2 y = 0$

So, Auxiliary equation is $(4D^2+3)^2=0 \implies D = \pm \frac{\sqrt{3}}{2}i, \pm \frac{\sqrt{3}}{2}i$ .

Hence, $y(x) = ( c_1 +c_2 x) cos(\frac{\sqrt{3}}{2} x) + (c_3 +c_4 x) sin(\frac{\sqrt{3}}{2} x)$ .

2) Given $y'' − 10y' + 26y = 0$

$\implies (D^2-10D+26)y = 0$ where $D = \frac{d}{dx}$ .

So, Auxiliary equation is $D^2-10D+26=0$

$\implies D = \frac{10\pm \sqrt{100-104}}{2} = 5 \pm i$ .

Hence, $y = e^{5x}(c_1cos(x)+c_2 sin(x))$