Question

Differential Equations

Answer 1

The differential equation

`ay'' + by' + c = 0`

has characteristic equation

`ar^2 + br + c = 0`

with roots
`r = (-b\pm√(b^2-4ac))/(2a) = (-b\pm√(D))/(2a)`.

• If
`D>0`, the roots are real and distinct, and the general solution is

`y = C_1 e^(r_1x) + C_2 e^(r_2x)`

• If
`D=0`, there is a repeated root and the general solution is

`y = C_1 e^(rx) + C_2 x e^(rx)`

• If
`D<0`, the roots are a complex conjugate pair
`r=\alpha\pm\beta i`, and the general solution is

`y = C_1 e^((\alpha+\beta i)x) + C_2 e^((\alpha-\beta i)x)`

which, by Euler's identity, can be expressed as

`y = C_1 e^(\alpha x) \cos(\beta x) + C_2 e^(\alpha x) \sin(\beta x)`

The solution curve in plot (A) has a somewhat periodic nature to it, so
`\boxed{D < 0}`. The plot suggests that
`y` will oscillate between -∞ and ∞ as
`x\to\infty`, which tells us
`\alpha>0` (otherwise, if
`\alpha=0` the curve would be a simple bounded sine wave, or if
`\alpha<0` the curve would still oscillate but converge to 0). Since
`\alpha` is the real part of the characteristic root, and we assume
`a>0`, we have

`\alpha = -\frac b{2a} > 0 \implies -b > 0 \implies \boxed{b < 0}`

Since
`D=b^2-4ac<0`, we have

`b^2 < 4ac \implies c > (b^2)/(4a) \implies \boxed{c>0}`

The solution curve in plot (B) is not periodic, so
`D\ge0`. For
`x` near 0, the exponential terms behave like constants (i.e.
`e^(rx)\to1`). This means that

• if
`D>0`, for some small neighborhood around
`x=0`, the curve is approximately constant,

`y = C_1 e^(r_1x) + C_2 e^(r_2x) \approx C_1 + C_2`

• if
`\boxed{D=0}`, for some small neighborhood around
`x=0`, the curve is approximately linear,

`y = C_1 e^(rx) + C_2 x e^(rx) \approx C_1 + C_2 x`

Since
`D=b^2-4ac=0`, it follows that

`b^2=4ac \implies c = (b^2)/(4a) \implies \boxed{c>0}`

As
`x\to\infty`, we see
`y\to-\infty` which means the characteristic root is positive (otherwise we would have
`y\to0`), and in turn

`r = -\frac b{2a} > 0 \implies -b > 0 \implies \boxed{b < 0}`