Respuesta :
Answer:
The general solution of second order homogeneous differential equation is [tex]y(x)=c_1e^{6x}+c_2e^{-5x}[/tex]
Step-by-step explanation:
To find the general solution of this second order homogeneous differential equation [tex]\frac{d^2y}{dx^2}-\frac{dy}{dx}-30y=0[/tex] we are going to use this Theorem:
Given the differential equation [tex]a \ddot{y}+b\dot{y}+cy =0, a\neq 0[/tex], consider the quadratic polynomial [tex]ax^2+bx+c[/tex], called the characteristic polynomial. Using the quadratic formula, this polynomial always has one or two roots, call them [tex]r[/tex] and [tex]s[/tex]. The general solution of the differential equation is:
(a) [tex]\ds y=Ae^{rt}+Be^{st}[/tex] if the roots [tex]r[/tex] and [tex]s[/tex] are real numbers and [tex]r\not=s[/tex].
(b) [tex]\ds y=Ae^{rt}+Bte^{rt}[/tex], if [tex]r=s[/tex] is real.
(c) [tex]\ds y=A\cos(\beta t)e^{\alpha t}+B\sin(\beta t)e^{\alpha t}[/tex], if the roots [tex]r[/tex] and [tex]s[/tex] are complex numbers [tex]\alpha+\beta i[/tex] and [tex]\alpha-\beta i[/tex]
Applying the above Theorem we have:
[tex]\mathrm{Substitute\quad }\frac{d^2y}{dx^2},\:\frac{dy}{dx}\mathrm{\:with\:}\ddot{y},\dot{y}[/tex]
[tex]\ddot{y}-\dot{y}-30y=0[/tex]
The characteristic polynomial is [tex]x^2-x-30[/tex] and we find the roots as follows:
[tex]\mathrm{Break\:the\:expression\:into\:groups}[/tex]
[tex]\left(x^2+5x\right)+\left(-6x-30\right)[/tex]
[tex]\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2+5x\mathrm{:\quad }x\left(x+5\right)[/tex]
[tex]\mathrm{Factor\:out\:}-6\mathrm{\:from\:}-6x-30\mathrm{:\quad }-6\left(x+5\right)[/tex]
[tex]\mathrm{Factor\:out\:common\:term\:}x+5[/tex]
[tex]\left(x+5\right)\left(x-6\right)[/tex]
The roots of characteristic polynomial are [tex]r=-5[/tex] and [tex]s=6[/tex]
Therefore the general solution of second order homogeneous differential equation is [tex]y(x)=c_1e^{6x}+c_2e^{-5x}[/tex]
