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

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

Explanation:

To find the general solution of this second order homogeneous differential equation
`(d^2y)/(dx^2)-(dy)/(dx)-30y=0` we are going to use this Theorem:

Given the differential equation
`a \ddot{y}+b\dot{y}+cy =0, a\\eq 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\\ot=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 }(d^2y)/(dx^2),\:(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)`