Linear Differential Equation with Variable Coefficients

Exercise 2.6

1. Solve the following Euler's equation. provided $x > 0$.

(a) $\ x^{2}y^{\prime\prime} + 4xy^{\prime} + 2y = 0$
(b) $\ x^{2}y^{\prime\prime} + xy^{\prime} + 9y = 0$
(c) $\ x^{2}y^{\prime\prime} - xy^{\prime} - 3y = x^{2}\log{x}$
(d) $\ x^{3}y^{\prime\prime\prime} + x^{2}y^{\prime\prime} -2xy^{\prime} + 2y = x^{3}$
(e) $\ x^{2}y^{\prime\prime\prime} + 5xy^{\prime\prime} + 4y^{\prime} = 0$

Answer
1. (a) Set $x = e^{t}, y = x^{\lambda}$. Then the indicial equation is

$$\lambda(\lambda - 1) + 4 \lambda + 2 = \lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0$$

Thus, we have $\lambda = -1, -2$. Note that this indicial equation is the characteristic equation of the following differential equation.

$$\frac{d^{2}y}{dt^{2}} + 3\frac{dy}{dt} + 2y = 0$$

Therefore, the general solution is

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

(b) Set $x = e^{t}, y = x^{\lambda}$. Then the indicial equation is given by

$$\lambda(\lambda - 1) + \lambda + 9 = \lambda^2 + 9 = 0$$

Thus, we have $\lambda = \pm 3i$. Note that this indicial equation is the characteristic equation of the following equation.

$$\frac{d^{2}y}{dt^{2}} + 9y = 0$$

Therefore, the general solution is given by

$$y = c_{1}\cos{3t} + c_{2}\sin{3t} = c_{1}\cos{(3\log{x})} + c_{2}\sin{(3\log{x})} \ \$$

(c) Set $x = e^{t}, y = x^{\lambda}$. Then the indicial equation is given by

$$\lambda(\lambda - 1) - \lambda - 3 = \lambda^2 - 2\lambda - 3 = (\lambda + 1)(\lambda - 3) = 0$$

Thus, we have $\lambda = -1, 3$. Note that this indicial equation is the characteristic equation of the following equation.

$$\frac{d^{2}y}{dt^{2}} - 2\frac{dy}{dt} - 3y = t e^{2t} \ \ *$$

Thus the complementary function $y_{c}$ is

$$y_{c} = c_{1}e^{-t} + c_{2}e^{3t}$$

We find the particular solution. Since $(D - 2)^{2} te^{2t} = 0$, we have

$$(D + 1)(D - 3)(D -2)^{2} y = 0$$

which implies the solutions are $\{e^{-t}, e^{3t}, e^{2t}, te^{2t}\}$. But $\{e^{-t}, e^{3t}\}$ are already used in the $y_c$. So, we omit these solutions from $y_p$. Then we have

$$y_{p} = Ate^{2t} + Be^{2t}$$

Substitute this into *. Then

$$\begin{array}{ll} -3(y_{p} & = Ate^{2t} + Be^{2t})\\ -2(y_{p... ...Be^{2t}}\\ te^{2t} & = (2A - 3B)e^{2t} -3Ate^{2t} \end{array}$$

Thus we have $A = -1/3, B = - 2/9$ and ，

$$y_{p} = -\frac{1}{3}te^{2t} - \frac{2}{9} e^{2t}$$

Therefore, the general solution is

$$y = c_{1}e^{-t} + c_{2}e^{3t} - (\frac{1}{3}te^{2t} + \frac{2}{9} e^{2t})$$

$$= c_{1}x^{-1} + c_{2}x^{3} - (\frac{1}{3}x^{2}\log{x} + \frac{2}{9}x^{2}) \ \$$

(d) Set $x = e^{t}, y = x^{\lambda}$. Then the indicial equation is given by

$$\lambda(\lambda - 1)(\lambda - 2) + \lambda(\lambda - 1) - 2\lambda + 2 = 0$$

Simplifying to get

$$\lambda(\lambda - 1)(\lambda - 2) + \lambda(\lambda - 1) - 2\lamb... ...a + 2) + \lambda^{2} - 3\lambda + 2 = (\lambda + 1)(\lambda -1)(\lambda -2) = 0$$

Thus, $\lambda = -1, 1, 2$. This indicial equation is the characteristic equation of the following differential equation.

$$(D + 1)(D - 1)(D - 2)y = e^{3t} \ \ \ \ {*}$$

Thus the complementary function $y_{c}$ is given by

$$y_{c} = c_{1}e^{-t} + c_{2}e^{t} + c_{3} e^{2t}$$

Now we find the particular solution using the variation of parameter. Since $(D - 3) e^{3t} = 0$, we have

$$(D + 1)(D - 1)(D -2)(D - 3) y = 0$$

Solve this for $y$ to get $\{e^{-t}, e^{t}, e^{2t}, e^{3t}\}$. But $\{e^{-t}, e^{t}, e^{2t}\}$ are already used in $y_c$. So, we omit from $y_p$. Then we have

$$y_{p} = Ae^{3t}$$

Substitute this into *.

$$(D^3 - 2D^2 - D + 2)Ae^{3t} = e^{3t}(2A - 3A - 18A + 27A) = e^{3t} (8A) = e^{3t}$$

Then $A = 1/8$ and

$$y_{p} = \frac{1}{8}e^{3t}$$

Therefore, the general solution is given by

$$y = c_{1}e^{-t} + c_{2}e^{t} + c_{3}e^{2t} + \frac{1}{8}e^{3t}$$

$$= c_{1}x^{-1} + c_{2}x + c_{3}x^{2} + \frac{1}{8}x^{3} \ \$$

(e) Given equation is not Euler's equation. But if we multiply $x$ to both sides. Then it becomes Euler's equation.

$$x^3 y^{\prime\prime\prime} + 5x^2 y^{\prime\prime} + 4xy^{\prime} = 0$$

Now let $x = e^{t}, y = x^{\lambda}$. Then the indicial equation is

$$\lambda(\lambda - 1)(\lambda - 2) + 5\lambda(\lambda - 1) + 4\lambda = 0$$

Simplify this equation. We get

$$\lambda^{3} + 2\lambda ^{2} + \lambda = \lambda(\lambda^2 + 2\lambda + 1) = \lambda(\lambda + 1)^2 = 0$$

Thus we have $\lambda = 0, -1, -1$. Not that this indicial equation is th echaracteristic equation of the following differential equation.

$$(D^3 + 2D^2 + D)y = 0$$

Therefore, the general solution is

$$y = c_{1} + c_{2}e^{-t} + c_{3}te^{-t} = c_{1} + c_{2}x^{-1} + c_{3}x^{-1}\log{x} \ \$$