Q31E

Question

Reduction of Order. If a nontrivial solution f(x) is known for the homogeneous equation

, $y^{(n)} + p_{1} (x) y^{(n - 1)} + . .. + p_{n} (x) y = 0$

the substitution $y (x) = v (x) f (x)$ can be used to reduce the order of the equation for second-order equations. By completing the following steps, demonstrate the method for the third-order equation

(35) $y''' - 2 y'' - 5 y' + 6 y = 0$

given that $f (x) = e^{x}$ is a solution.

(a) Set $y (x) = v (x) e^{x}$ and compute y′, y″, and y‴.

(b) Substitute your expressions from (a) into (35) to obtain a second-order equation in. $w = v'$

(c) Solve the second-order equation in part (b) for w and integrate to find v. Determine two linearly independent choices for v, say, $v_{1}$ and $v_{2}$ .

(d) By part (c), the functions $y_{1} (x) = v_{1} (x) e^{x}$ and $y_{2} (x) = v_{2} (x) e^{x}$ are two solutions to (35). Verify that the three solutions $e^{x}, y_{1} (x)$ , and $y_{2} (x)$ are linearly independent on $(- \infty, \infty)$

Step-by-Step Solution

Verified

Answer

(a) The value of y′, y″, and y‴ is,

$\begin{matrix} y' = (v + v') e^{x} \\ y'' = (v + 2 v' + v'') e^{x} \\ y''' = (v + 3 v' + 3 v'' + v''') e^{x} \end{matrix}$

(b) A second-order equation is,

$w'' + w' - 6 w = 0$

(d) $w \neq 0$ and now we can say that $e^{x}, y_{1}, y_{2}$ are linearly independent solutions.

1Step 1: About Reduction of Order;

Now going to take a brief detour and look at solutions to non-constant coefficient, second order differential equations of the form.

$p (t) y'' + q (t) y' + r (t) y = 0$

In general, finding solutions to these kinds of differential equations can be much more difficult than finding solutions to constant coefficient differential equations. This method is called reduction of order.

2(a) Step 2: Firstly, using the given function f ( x ) = e x ,

Given function,

$f (x) = e^{x}$ is a solution to

$y''' - 2 y'' - 5 y' + 6 y = 0 ...... (1)$

And

$y (x) = v (x) f (x) = v (x) e^{x}$

Now find the derivative of y for equation (1),

$\begin{matrix} y = {ve}^{x} \\ y' = {ve}^{x} + v' e^{x} \\ y'' = {ve}^{x} + v' e^{x} + v'' e^{x} + v' e^{x} \\ y'' = {ve}^{x} + 2 v' e^{x} + v'' e^{x} \\ y''' = {ve}^{x} + v' e^{x} + 2 v'' e^{x} + 2 v' e^{x} + v''' e^{x} + v'' e^{x} \\ y''' = (v + 3 v' + 3 v'' + v''') e^{x} \end{matrix}$

Hence, the value of y′, y″, and y‴ is,

$\begin{matrix} y' = (v + v') e^{x} \\ y'' = (v + 2 v' + v'') e^{x} \\ y''' = (v + 3 v' + 3 v'' + v''') e^{x} \end{matrix}$

3(b) Step 3: Obtain a second-order equation in . w = v '

Substitute all values in the equation (1),

$\begin{matrix} y''' - 2 y'' - 5 y' + 6 y = 0 \\ {ve}^{x} + 3 v' e^{x} + 3 v'' e^{x} + v''' e^{x} - 2 ({ve}^{x} + 2 v' e^{x} + v'' e^{x}) - 5 ({ve}^{x} + v' e^{x}) + 6 {ve}^{x} = 0 \\ - 6 v' e^{x} + v'' e^{x} + v''' e^{x} = 0 \\ - 6 v' + v'' + v''' = 0 \end{matrix}$

Use the value $w = v'$ in the above expression,

Hence, A second-order equation is,

$w'' + w' - 6 w = 0$

4(c) Step 4: Solve the second-order equation in part (b) for w,

Solve the above equation for w,

$\begin{matrix} (D^{2} + D - 6) w = 0 \\ D = - 3, 2 \end{matrix}$

The solution of w is,

$\begin{matrix} w (x) = {Ae}^{- 3 x} + {Be}^{2 x} \\ v' = {Ae}^{- 3 x} + {Be}^{2 x} \end{matrix}$

Integrating both sides with respect to x,

$\begin{matrix} \int v' = \int ({Ae}^{- 3 x} + {Be}^{2 x}) dx \\ v = \frac{{Ae}^{- 3 x}}{- 3} + \frac{{Be}^{2 x}}{2} + C \\ v = v_{1} + v_{2} \\ v_{1} = e^{- 3 x} \\ v_{2} = e^{2 x} \end{matrix}$

Hence, two linearly independent is, . $v_{1} = e^{- 3 x}, v_{2} = e^{2 x}$

5(d) Step 5: Verify that the three solutions, e x ,   y 1 ( x ) and y 2 ( x ) are linearly independent on. ( - ∞ ,   ∞ )

We have,

$\begin{array}{l} y_{i} = v_{i} f \\ y_{1} = v_{1} e^{x} = e^{- 2 x} \\ y_{2} = v_{2} e^{x} = e^{3 x} \end{array}$

To Verify, find the derivative of $y_{1}$ and $y_{2}$

$\begin{array}{l} y_{1}' = - 2 e^{- 2 x}, y_{1}'' = 4 e^{- 2 x}, y_{1}''' = - 8 e^{- 2 x} \\ y_{2}' = 3 e^{3 x}, y_{2}'' = 9 e^{3 x}, y_{2}''' = 27 e^{3 x} \end{array}$

Now,

$\begin{matrix} y_{1}''' - 2 y_{1}'' - 5 y_{1}' + 6 y_{1} = y_{2}''' - 2 y_{2}'' - 5 y_{2}' + 6 y_{2} \\ - 8 e^{- 2 x} - 2 (4 e^{- 2 x}) - 5 (- 2 e^{- 2 x}) + 6 (e^{- 2 x}) = 27 e^{3 x} - 2 (9 e^{3 x}) - 5 (3 e^{3 x}) + 6 (e^{3 x}) \\ - 8 e^{- 2 x} - 8 e^{- 2 x} + 10 e^{- 2 x} + 6 e^{- 2 x} = 27 e^{3 x} - 18 e^{3 x} - 15 e^{3 x} + 6 e^{3 x} \\ 0 = 0 \end{matrix}$

Using the Wronskian,

$\begin{matrix} w = | \begin{matrix} e^{x} & e^{- 2 x} & e^{3 x} \\ e^{x} & - 2 e^{- 2 x} & 3 e^{3 x} \\ e^{x} & 4 e^{- 2 x} & 9 e^{3 x} \end{matrix} | \\ = e^{x} (- 18 e^{x} - 12 e^{x}) - e^{- 2 x} (9 e^{4 x} - 3 e^{4 x}) + e^{3 x} (4 e^{- x} + 2 e^{- x}) \\ = - 30 e^{2 x} \\ \neq 0 \end{matrix}$

Hence, $w \neq 0$ and now we can say that $e^{x}, y_{1}, y_{2}$ are linearly independent solutions.

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