Solution of Linear Differential Equations

Exercise 2.1

1.
The following differential equations have the solution of the form $e^{mx}$. Find the n independent solutions. Then show the general solution. Also, show that the solutons are linearly independent by using Wronski's determinant.

(a) $\ y^{\prime\prime\prime} + y^{\prime\prime} -10y^{\prime} + 8y = 0$
(b) $\ y^{\prime\prime} + 4y^{\prime} + 3y = 0$
(c) $\ y^{\prime\prime} - y^{\prime} = 0$
(d) $\ y^{\prime\prime\prime} - 8y^{\prime\prime} + 7y^{\prime} = 0$
2. The following differential equations have the solution of the form $\cos{mx},\sin{mx}$. Find the general solution..
(a) $y^{\prime\prime} + 4y = 0 \htmlref{(b)}{enshu:2-1-1-b} y^{(4)} + 4y^{\prime\prime} + 3y = 0$
3. The following differential equations have the solution of the form $x^{m}$. FInd the general solution.
(a) $x^{2}y^{\prime\prime} + xy^{\prime} - 4y = 0 \htmlref{(b)}{enshu:2-1-3-b} x^{2}y^{\prime\prime} - xy^{\prime} - 3y = 0$

Answer
1. (a) Let $L(y) = y^{\prime\prime\prime} + y^{\prime\prime} -10y^{\prime} + 8y$. Then

$$L(e^{mx}) = m^{3}e^{mx} + m^{2}e^{mx} - 10me^{mx} + 8e^{mx}$$

$$= e^{mx}(m^3 + m^2 - 10m + 8)$$

$$= e^{mx}(m-1)(m^2 + 2m - 8) = e^{mx}(m-1)(m-2)(m+4)$$

Then, for $m = -4, 1, 2$， $L(e^{mx}) = 0$. Thus, $e^{-4x},e^{x},e^{2x}$ are solutions and

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

now to show these solutions are linearly independent, we need to show Wronski's determinant is not 0. Thus,

$$W(e^{-4x},e^{x},e^{2x}) = \left \vert \begin{array}{ccc} e^{-4x}... ... \\ -4 & 1 & 2 \\ 16 & 1 & 4 \end{array} \right \vert = 30e^{-x} \neq 0 \$$

(b) Let $L(y) = y^{\prime\prime} + 4y^{\prime} + 3y$. Then

$$L(e^{mx}) = m^{2}e^{mx} + 4me^{mx} + 3e^{mx}$$

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

$$= e^{mx}(m+1)(m+3)$$

Then for $m = -3, -1$， $L(e^{mx}) = 0$. Thus, $e^{-3x},e^{-x}$ are solutions and the general solution is

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

To show that these solutions are linearly independent, Wronski's determinant is not 0. Thus,

$$W(e^{-3x},e^{-x}) = \left \vert \begin{array}{cc} e^{-3x} & e^{-x} \\ -3e^{-3x} & -e^{-x} \end{array} \right \vert = 2e^{-4x} \neq 0 \$$

(c) Let $L(y) = y^{\prime\prime} - y^{\prime}$. Then

$$L(e^{mx}) = m^{2}e^{mx} - me^{mx}$$

$$= e^{mx}(m^2 - m)$$

$$= e^{mx}(m(m - 1))$$

Thus, for $m = 0 , 1$， $L(e^{mx}) = 0$. Then $1,e^{x}$ are solutions and the general solution is

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

To show these solutions are linearly independet, it is enough to show that Wronski's determinant is not 0.

$$W(1,e^{x}) = \left \vert \begin{array}{cc} 1 & e^{x} \\ 0 & e^{x} \end{array} \right \vert = e^{x} \neq 0 \$$

(d) Let $L(y) = y^{\prime\prime\prime} - 8y^{\prime\prime} + 7y^{\prime}$. Then

$$L(e^{mx}) = m^{3}e^{mx} - 8m^{2}e^{mx} + 7me^{mx}$$

$$= e^{mx}(m^3 - 8m^{2} + 7m)$$

$$= e^{mx}(m(m^2 - 8m + 7)) = e^{mx}(m(m-1)(m-7))$$

Thus for $m = 0, 1, 7$， $L(e^{mx}) = 0$. Then $1,e^{x},e^{7x}$ are solutions and the general solution is

$$y = c_{1} + c_{2}e^{x} + c_{3}e^{7x}$$

To show these solutions are linearly independent, it is enough to show Wronski's determinant is not 0.

$$W(1,e^{x},e^{7x}) = \left \vert \begin{array}{ccc} 1 & e^{x} & e... ...e^{7x}\\ 0 & e^{x} & 49 e^{7x} \end{array} \right \vert = 42 e^{8x} \neq 0 \$$

2. (a) Let $L(y) = y^{\prime\prime} + 4y$. Then

$$L(\cos{mx}) = -m^{2}\cos{mx} + 4\cos{mx}$$

$$= \cos{mx}(4 - m^2)$$

$$= \cos{mx}(2 + m)(2 - m)$$

Thus for $m = -2, 2$， $L(\cos{mx}) = 0$. Thus, $\cos{(-2x)} = \cos{2x}$ is a solution.

$$L(\sin{mx}) = -m^{2}\sin{mx} + 4\sin{mx}$$

$$= \sin{mx}(4 - m^2)$$

$$= \sin{mx}(2 + m)(2 - m)$$

Thus for $m = -2, 2$， $L(\sin{mx}) = 0$. Thusm $\sin{-2x} = -\sin{2x}$ is also a solution. Note that for $\cos{2x}$ and $\sin{2x}$, we have

$$W(\cos{2x},\sin{2x}) = \left \vert \begin{array}{cc} \cos{2x} & \sin{2x} \\ -2\sin{2x} & 2\cos{2x} \end{array} \right \vert = 2$$

Thus, they are linearly independent and the general solution is

$$y = c_{1}\cos{2x} + c_{2}\sin{2x} \ \$$

(b) Let $L(y) = y^{(4)} + 4y^{\prime\prime} + 3y$. Then

$$L(\cos{mx}) = m^{4}\cos{mx} - 4m^{2}\cos{mx} + 3\cos{mx}$$

$$= \cos{mx}(m^4 - 4m^2 + 3)$$

$$= \cos{mx}(m^2 - 1)(m^2 - 3)$$

Thus for $m = \pm 1, \pm \sqrt{3}$， $L(\cos{mx}) = 0$. Then $\cos{x}, \cos{\sqrt{3}x}$ are solutions.

$$L(\sin{mx}) = m^4 \sin{mx} - 4m^{2}\sin{mx} + 3\sin{mx}$$

$$= \sin{mx}(m^4 - 4m^2 + 3)$$

$$= \sin{mx}(m^2 - 1)(m^2 - 3)$$

Then for $m = \pm 1, \pm \sqrt{3}$， $L(\sin{mx}) = 0$. Thus, $\sin{x}, \sin{\sqrt{3}x}$ are also solutions. Now we check to see $\cos{x}, \cos{\sqrt{3}x}, \sin{x}, \sin{\sqrt{3}x}$ are linearly independent.

$$W = \left \vert \begin{array}{cccc} \cos{x} & \cos{\sqrt{3}x} & \sin{... ...n{\sqrt{3}x} & -\cos{x} & -3\sqrt{3}\cos{\sqrt{3}x} \\ \end{array}\right \vert$$

$$= -4\sqrt{3}$$

Thus, it is linearly indenpendent. Therefore, the general solution is

$$y = c_{1}\cos{x} + c_{2}\sin{x} + c_{3}\cos{\sqrt{3}x} + c_{4}\sin{\sqrt{3}x} \ \$$

3. (a) Let $L(y) = x^{2}y^{\prime\prime} + x y^{\prime} - 4 y$.

$$L(x^{m}) = x^{2}m(m-1)x^{m-2} + x mx^{m-1} - 4x^{m}$$

$$= x^{m}(m^2 - m + m - 4) = x^{m}(m^2 - 4)$$

$$= x^{m}(m + 2)(m - 2)$$

Then for $m = \pm 2$， $L(x^{m}) = 0$. Thus, $x^{-2}, x^{2}$ are solutions. Note that $x^{-2}, x^{2}$ satisfies

$$W(x^{-2}, x^{2}) = \left \vert \begin{array}{cc} x^{-2} & x^{2} \\ -2x^{-1} & 2x \end{array} \right \vert = 4x^{-1}$$

Thus they are linearly independent. Therefore, the general solution is

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

(b) Let $L(y) = x^{2}y^{\prime\prime} - x y^{\prime} - 3y$.

$$L(x^{m}) = x^{2}m(m-1)x^{m-2} - x mx^{m-1} - 3x^{m}$$

$$= x^{m}(m^2 - m - m - 3) = x^{m}(m^2 - 2m - 3)$$

$$= x^{m}(m + 1)(m - 3)$$

Then for $m = -1, 3$， $L(x^{m}) = 0$. Thus, $x^{-1}, x^{3}$ are solutions. Note that $x^{-1}, x^{3}$ satisfies

$$W(x^{-1}, x^{3}) = \left \vert \begin{array}{cc} x^{-1} & x^{3} \\ -x^{-2} & 3x^{2} \end{array} \right \vert = 4x$$

Thus, it is linearly independent and the general solution is

$$y = c_{1}x^{-1} + c_{2}x^{3} \ \$$