Higer Order Homogeneous Linear Differential Equations

Exercise 2.3

1. Find the fundamental solution of the following differential equations.

(a) $\ y^{\prime\prime} - 9y = 0$
(b) $\ y^{\prime\prime\prime} + y = 0$
(c) $\ y^{\prime\prime\prime} - 3y^{\prime} - 2y = 0$
(d) $\ y^{(5)} + 18y^{\prime\prime\prime} + 81y^{\prime} = 0$

2. Find the general solution of the following differential equations.
(a) 4th order homogeneous differential equation with the roots of the characteristic equation are $m = -2,1,1,1$
(b) 6th order homogeneous differential equation with the roots of the characteristic equation are $m = 0,0,-1\pm2i,-1\pm2i$
3. Solve the following initial value problems.
(a) $y^{\prime\prime} - y = 0, y(0) = 1, y^{\prime}(0) = 1$
(b) $y^{\prime\prime} -6y^{\prime} +9y = 0, y(0) = 1, y^{\prime}(0) = 2$
(c) $y^{\prime\prime\prime} + 7y^{\prime\prime} + 19y^{\prime} + 13y = 0, y(0) = 0, y^{\prime}(0) = 2, y^{\prime\prime}(0) = -12$
(d) $y^{(4)} + 2y^{\prime\prime\prime} + 10y^{\prime\prime} = 0, y(0) = 5, y^{\prime}(0) = -3, y^{\prime\prime}(0) = 0, y^{\prime\prime\prime}(0) = 0$

Answer
1. (a) Let $y = e^{mx}$. Then we have the characteristic equation $m^2 - 9 = 0$. Thus, the characteristic roots are $m = 3, 3$. Therefore, the fundamental solutions are

$$\{e^{3x}, xe^{3x} \}$$

(b) Let $y = e^{mx}$. Then we have the characteristic equation $m^3 + 1 = 0$.

$$m^3 + 1 = (m + 1)(m^2 - m + 1)$$

Then the characteristic roots are $m = -1, \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{3}i}{2}$. Thus, the fundamental solutions are

$$\{ e^{-x}, e^{\frac{x}{2}}\cos{(\frac{\sqrt{3}x}{2})}, e^{\frac{x}{2}}\sin{(\frac{\sqrt{3}x}{2})} \}$$

(c) Let $y = e^{mx}$. Then we have the characterisitc equation $m^3 - 3m - 2 = 0$.

$$m^3 - 3m - 2 = (m + 1)(m^2 - m - 2) = ( m + 1)(m + 1)(m - 2)$$

Then the characteristic roots are $m = -1, -1, 2$. Therefore, the fundamental solutions are

$$\{ e^{-x}, xe^{-x}, e^{2x} \}$$

(d) Let $y = e^{mx}$. Then we have the characteristic equation $m^5 + 18 m^3 + 81m = 0$.

$$m^5 + 18 m^3 + 81m = m(m^4 + 18m^2 + 81) = m( m^2 + 9)^{2}$$

Then the characteristic roots are $m = 0, \pm 3i, \pm 3i$. Therefore, the fundamental solutions are

$$\{ 1, \cos{3x}, x\cos{3x}, \sin{3x}, x\sin{3x} \}$$

2. (a) Let the roots of the characteristic equation be $m = -2,1,1,1$. Then the fundamental solutions are

$$\{e^{-2x}, e^{x}, xe^{x}, x^2 e^{x} \}$$

Therefore, the general solution is given by

$$y = c_{1}e^{-2x} + (c_{2} + c_{3}x + c_{4}x^2)e^{x} \ \$$

(b) Let the roots of the characteristic equation be $m = 0,0,-1 + 2i, -1 + 2i$. Then the fundamental solutions are

$$\{ 1, x, e^{-x}\cos{2x}, e^{-x}\sin{2x}, xe^{-x}\cos{2x}, xe^{-x}\sin{2x} \}$$

Therefore, the general solution is given by

$$y = c_{1} + c_{2}x + e^{-x}[(c_{3} + c_{4})\cos{2x} + (c_{5} + c_{6}x)\sin{2x}] \ \$$

3. (a) Let $y = e^{mx}$. Then we have the characteristic equation $m^2 - 1 = 0$.

$$m^2 - 1 = (m + 1)(m-1)$$

Then the characteristic roots are $m = -1, 1$. Thus, the fundamental solutions are

$$\{ e^{-x}, e^{x} \}$$

Therefore, the general solution is given by

$$y = c_{1}e^{-x} + c_{2}e^{x}$$

Now using the initial conditions $y(0) = 1, y^{\prime}(0) = 1$. Then

$$1 = y(0) = c_{1} + c_{2}$$

$$1 = y^{\prime}(0) = -c_{1} + c_{2}$$

Solve this system of linear equations. Then $c_{1} = 0, c_{2} = 1$. Therefore,

$$y = e^{x} \ \$$

(b) Let $y = e^{mx}$. Then we have the characteristic equation $m^2 - 6m + 9 = 0$.

$$m^2 - 6m + 9 = (m -3)^2$$

Then the characteristic roots are $m = 3, 3$. Thus, the fundamental solutions are

$$\{e^{3x}, xe^{3x} \}$$

Therefore, the general solution is given by

$$y = c_{1}e^{3x} + c_{2}xe^{3x}$$

Now using the initial conditions $y(0) = 1, y^{\prime}(0) = 2$, we have

$$1 = y(0) = c_{1}$$

$$2 = y^{\prime}(0) = 3c_{1} + c_{2}$$

Solving this system of linear equations, we have $c_{1} = 1, c_{2} = -1$. Therefore,

$$y = (1 - x)e^{3x} \ \$$

(c) Let $y = e^{mx}$. Then we have the characteristic equation $m^3 + 7m^2 + 19m + 13 = 0$.

$$m^3 + 7m^2 + 19m + 13 = (m + 1)(m^2 + 6m + 13)$$

The characteristic roots are $m = -1$ and $m = \ -3 \pm \sqrt{3^2 - 13} = -3 \pm \sqrt{-4} = -3 \pm 2i$. Thus, the general solutions are

$$\{ e^{-x}, e^{-3x}\cos{2x}, e^{3x}\sin{2x} \}$$

The general solution is given by

$$y = c_{1}e^{-x} + e^{-3x}[c_{2}\cos{2x} + c_{3}\sin{2x}]$$

Using the initial conditions $y(0) = 0, y^{\prime}(0) = 2, y^{\prime\prime}(0) = -12$ and

$$y^{\prime} = -c_{1}e^{-x} -3e^{-3x}[c_{2}\cos{2x} + c_{3}\sin{2x}]$$

$$+ e^{-3x}[-2c_{2}\sin{2x} + 2c_{3}\cos{2x}]$$

$$y^{\prime\prime} = c_{1}e^{-x} + 9e^{-3x}[c_{2}\cos{2x} + c_{3}\sin{2x}]$$

$$- 6e^{-3x}[-2c_{2}\sin{2x} + 2c_{3}\cos{2x}] + e^{-3x}[-4c_{2}\cos{2x} - 4c_{3}\sin{2x}]$$

Then we have

$$= y(0) = c_{1} + c_{2}$$ 0 (2.1)

$$2 = y^{\prime}(0) = -c_{1} -3 c_{2} + 2c_{3} \$$ (2.2)

$$-12 = y^{\prime\prime}(0) = c_{1} + 5c_{2} +12c_{3} \$$ (2.3)

Solving this system of linear equations, we have $2 = -2c_{2} + 2c_{3}, \ -12 = 6c_{2} + 12c_{3}$ implies $c_{1} = 4/3, c_{2} = -4/3, c_{3} = -1/3$. Therefore,

$$y = \frac{4}{3}e^{-x} - e^{-3x}[\frac{4}{3}\cos{2x} + \frac{1}{3}\sin{2x}] \ \$$

(d) Let $y = e^{mx}$. Then we have the characteristic equation $m^4 + 2m^3 + 10m^2 = 0$.

$$m^4 + 2m^3 + 10m^2 = m^{2}(m^2 + 2m + 10)$$

Then the characteristic roots are $m = 0, 0$ and $m = -1 \pm \sqrt{1^2 - 10} = -1 \pm \sqrt{-9} = -1 \pm 3i$. Thus the fundamental solutions are

$$\{ 1, x, e^{-x}\cos{3x}, e^{-x} \sin{3x}\}$$

The general solution is given by

$$y = c_{1} + c_{2}x + e^{-x}[c_{3}\cos{3x} + c_{4}\sin{3x}]$$

Now using the initial condition $y(0) = 5, y^{\prime}(0) = -3, y^{\prime\prime}(0) = 0, y^{\prime\prime\prime}(0) = 0$ and

$$y^{\prime} = c_{2} - e^{-x}[c_{3}\cos{3x} + c_{4}\sin{3x}]$$

$$+ e^{-x}[-3c_{3}\sin{3x} + 3c_{4}\cos{3x}$$

$$y^{\prime\prime} = e^{-x}[c_{3}\cos{3x} + c_{4}\sin{3x}]$$

$$- 2e^{-x}[-3c_{3}\sin{3x} + 3c_{4}\cos{3x}] + e^{-x}[-9c_{3}\cos{3x} - 9c_{4}\sin{3x}]$$

$$y^{\prime\prime\prime} = - e^{-x}[c_{3}\cos{3x} + c_{4}\sin{3x}] + 3e^{-x}[-3c_{3}\sin{3x} + 3c_{4}\cos{3x}]$$

$$- 3e^{-x}[-9c_{3}\cos{3x} - 9c_{4}\sin{3x}] + e^{-x}[27c_{3}\sin{3x} -27c_{4}\cos{3x}]$$

Then

$$5 = y(0) = c_{1} + c_{3} \$$ (2.4)

$$-3 = y^{\prime}(0) = c_{2} - c_{3} + 3c_{4} \$$ (2.5)

$$= y^{\prime\prime}(0) = c_{3} - 6c_{4} - 9c_{3} \$$ 0 (2.6)

$$= y^{\prime\prime\prime}(0) = -c_{3} + 9c_{4} + 27 c_{3} - 27 c_{4}$$ 0 (2.7)

Solving this system of linear equations   we have $c_{1} = 5, c_{2} = -3, c_{3} = c_{4} = 0$. Thus,

$$y = 5 - 3x \ \$$