Question

Find the general solution to the homogeneous differential equation. d2ydx2−1dydx−30y=0 d2ydx2−1dydx−30y=0 Use c1c1 and c2c2 in your answer to denote arbitrary constants, and enter them as c1 and c2.

Answer 1

Answer:

The general solution of second order homogeneous differential equation is $y(x)=c_1e^{6x}+c_2e^{-5x}$

Step-by-step explanation:

To find the general solution of this second order homogeneous differential equation $\frac{d^2y}{dx^2}-\frac{dy}{dx}-30y=0$ we are going to use this Theorem:

Given the differential equation $a \ddot{y}+b\dot{y}+cy =0, a\neq 0$, consider the quadratic polynomial $ax^2+bx+c$, called the characteristic polynomial. Using the quadratic formula, this polynomial always has one or two roots, call them $r$ and $s$. The general solution of the differential equation is:

(a) $\ds y=Ae^{rt}+Be^{st}$ if the roots $r$ and $s$ are real numbers and $r\not=s$.

(b) $\ds y=Ae^{rt}+Bte^{rt}$, if $r=s$ is real.

(c) $\ds y=A\cos(\beta t)e^{\alpha t}+B\sin(\beta t)e^{\alpha t}$, if the roots $r$ and $s$ are complex numbers $\alpha+\beta i$ and $\alpha-\beta i$

Applying the above Theorem we have:

$\mathrm{Substitute\quad }\frac{d^2y}{dx^2},\:\frac{dy}{dx}\mathrm{\:with\:}\ddot{y},\dot{y}$

$\ddot{y}-\dot{y}-30y=0$

The characteristic polynomial is $x^2-x-30$ and we find the roots as follows:

$\mathrm{Break\:the\:expression\:into\:groups}$

$\left(x^2+5x\right)+\left(-6x-30\right)$

$\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2+5x\mathrm{:\quad }x\left(x+5\right)$

$\mathrm{Factor\:out\:}-6\mathrm{\:from\:}-6x-30\mathrm{:\quad }-6\left(x+5\right)$

$\mathrm{Factor\:out\:common\:term\:}x+5$

$\left(x+5\right)\left(x-6\right)$

The roots of characteristic polynomial are $r=-5$ and $s=6$

Therefore the general solution of second order homogeneous differential equation is $y(x)=c_1e^{6x}+c_2e^{-5x}$