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andre [41]
3 years ago
13

Find the general solution to the homogeneous differential equation. d2ydx2−1dydx−30y=0 d2ydx2−1dydx−30y=0 Use c1c1 and c2c2 in y

our answer to denote arbitrary constants, and enter them as c1 and c2.
Mathematics
1 answer:
lukranit [14]3 years ago
5 0

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:

<em>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</em><em> characteristic polynomial.</em><em> 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:</em>

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

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

<em>(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</em>

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}

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