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Expert-verified Found in: Page 327 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Reduction of Order. If a nontrivial solution f(x) is known for the homogeneous equation,${{\mathbf{y}}}^{\left(\mathbf{n}\right)}{\mathbf{+}}{{\mathbf{p}}}_{1}\left(\mathbf{x}\right){{\mathbf{y}}}^{\left(\mathbf{n}\mathbf{-}\mathbf{1}\right)}{\mathbf{+}}{.}{..}{\mathbf{+}}{{\mathbf{p}}}_{n}\left(\mathbf{x}\right){\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$the substitution ${\mathbf{y}}\left(\mathbf{x}\right){\mathbf{=}}{\mathbf{v}}\left(\mathbf{x}\right){\mathbf{f}}\left(\mathbf{x}\right)$ 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) ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{5}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{6}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$given that ${\mathbf{f}}\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{e}}}^{x}$ is a solution.(a) Set ${\mathbf{y}}\left(\mathbf{x}\right){\mathbf{=}}{\mathbf{v}}\left(\mathbf{x}\right){{\mathbf{e}}}^{x}$ and compute y′, y″, and y‴. (b) Substitute your expressions from (a) into (35) to obtain a second-order equation in. ${\mathbf{w}}{\mathbf{=}}{\mathbf{v}}{\mathbf{\text{'}}}$(c) Solve the second-order equation in part (b) for w and integrate to find v. Determine two linearly independent choices for v, say, ${\mathbf{v}}_{1}$and ${{\mathbf{v}}}_{2}$. (d) By part (c), the functions ${{\mathbf{y}}}_{1}\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{v}}}_{1}\left(\mathbf{x}\right){{\mathbf{e}}}^{x}$ and ${{\mathbf{y}}}_{2}\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{v}}}_{2}\left(\mathbf{x}\right){{\mathbf{e}}}^{x}$ are two solutions to (35). Verify that the three solutions ${{\mathbf{e}}}^{x}{\mathbf{,}}{\text{\hspace{0.17em}}}{{\mathbf{y}}}_{1}\left(\mathbf{x}\right)$, and ${{\mathbf{y}}}_{2}\left(\mathbf{x}\right)$ are linearly independent on $\left(\mathbf{-}\infty \mathbf{,}\text{\hspace{0.17em}}\infty \right)$

(a) The value of y′, y″, and y‴ is,

$\begin{array}{c}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\end{array}$

(b) A second-order equation is,

$\mathbf{w}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{w}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{w}\mathbf{=}\mathbf{0}$

(c) Two linearly independent is,${\mathbf{v}}_{1}\mathbf{=}{\mathbf{e}}^{-3x},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathbf{v}}_{2}\mathbf{=}{\mathbf{e}}^{2x}$

(d) $\mathbf{w}\ne \mathbf{0}$ and now we can say that ${\mathbf{e}}^{x},\text{\hspace{0.17em}\hspace{0.17em}}{\mathbf{y}}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathbf{y}}_{2}$are linearly independent solutions.

See the step by step solution

## Step 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.

$\mathbf{p}\left(\mathbf{t}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{q}\left(\mathbf{t}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{r}\left(\mathbf{t}\right)\mathbf{y}\mathbf{=}\mathbf{0}$

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.

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

Given function,

$\mathbf{f}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^{x}$ is a solution to

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(\mathbf{1}\right)$

And

$\mathbf{y}\left(\mathbf{x}\right)\mathbf{=}\mathbf{v}\left(\mathbf{x}\right)\mathbf{f}\left(\mathbf{x}\right)\mathbf{=}\mathbf{v}\left(\mathbf{x}\right){\mathbf{e}}^{x}$

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

$\begin{array}{c}\mathbf{y}\mathbf{=}{\mathbf{ve}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}{\mathbf{ve}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{ve}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{ve}}^{x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{ve}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\end{array}$

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

$\begin{array}{c}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\end{array}$

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

Substitute all values in the equation (1),

$\begin{array}{c}y\text{'}\text{'}\text{'}-2y\text{'}\text{'}-5y\text{'}+6y=0\\ {\mathbf{ve}}^{x}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{-}\mathbf{2}\left({\mathbf{ve}}^{x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\right)\mathbf{-}\mathbf{5}\left({\mathbf{ve}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\right)\mathbf{+}\mathbf{6}{\mathbf{ve}}^{x}\mathbf{=}\mathbf{0}\\ \mathbf{-}\mathbf{6}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{=}\mathbf{0}\\ -6v\text{'}+v\text{'}\text{'}+v\text{'}\text{'}\text{'}=0\end{array}$

Use the value $\mathbf{w}\mathbf{=}\mathbf{v}\mathbf{\text{'}}$in the above expression,

Hence, A second-order equation is,

${\mathbf{w}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{w}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{6}}{\mathbf{w}}{\mathbf{=}}{\mathbf{0}}$

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

Solve the above equation for w,

$\begin{array}{c}\left({\mathbf{D}}^{2}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{6}\right)\mathbf{w}\mathbf{=}\mathbf{0}\\ \mathbf{D}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\text{\hspace{0.17em}}\mathbf{2}\end{array}$

The solution of w is,

$\begin{array}{c}\mathbf{w}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}3\mathbf{x}}\mathbf{+}{\mathbf{Be}}^{2x}\\ \mathbf{v}\mathbf{\text{'}}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}3\mathbf{x}}\mathbf{+}{\mathbf{Be}}^{2x}\end{array}$

Integrating both sides with respect to x,

$\begin{array}{c}\int \mathbf{v}\mathbf{\text{'}}\mathbf{=}\int \left({\mathbf{Ae}}^{-3x}\mathbf{+}{\mathbf{Be}}^{2x}\right)\mathbf{dx}\\ \mathbf{v}\mathbf{=}\frac{{\mathbf{Ae}}^{-3x}}{-3}\mathbf{+}\frac{{\mathbf{Be}}^{2x}}{2}\mathbf{+}\mathbf{C}\\ \mathbf{v}\mathbf{=}{\mathbf{v}}_{1}\mathbf{+}{\mathbf{v}}_{2}\\ {\mathbf{v}}_{1}\mathbf{=}{\mathbf{e}}^{-3x}\\ {\mathbf{v}}_{2}\mathbf{=}{\mathbf{e}}^{2x}\end{array}$

Hence, two linearly independent is, .${{\mathbf{v}}}_{1}{\mathbf{=}}{{\mathbf{e}}}^{-3x}{,}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{{\mathbf{v}}}_{2}{\mathbf{=}}{{\mathbf{e}}}^{2x}$

## (d)Step 5: Verify that the three solutions,  ex, y1(x) and y2(x) are linearly independent on.(-∞, ∞)

We have,

$\begin{array}{l}{\mathbf{y}}_{i}\mathbf{=}{\mathbf{v}}_{i}\mathbf{f}\\ {\mathbf{y}}_{1}\mathbf{=}{\mathbf{v}}_{1}{\mathbf{e}}^{x}\mathbf{=}{\mathbf{e}}^{-2x}\\ {\mathbf{y}}_{2}\mathbf{=}{\mathbf{v}}_{2}{\mathbf{e}}^{x}\mathbf{=}{\mathbf{e}}^{3x}\end{array}$

To Verify, find the derivative of ${\mathbf{y}}_{1}$ and ${\mathbf{y}}_{2}$

$\begin{array}{l}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{e}}^{-2x}\mathbf{,}\text{\hspace{0.17em}}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{4}{\mathbf{e}}^{-2x}\mathbf{,}\text{\hspace{0.17em}}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{8}{\mathbf{e}}^{-2x}\\ {\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{=}\mathbf{3}{\mathbf{e}}^{3x}\mathbf{,}\text{\hspace{0.17em}}{\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{9}{\mathbf{e}}^{3x}\mathbf{,}\text{\hspace{0.17em}}{\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{27}{\mathbf{e}}^{3x}\end{array}$

Now,

$\begin{array}{c}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{+}\mathbf{6}{\mathbf{y}}_{1}\mathbf{=}{\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}{\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}{\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{+}\mathbf{6}{\mathbf{y}}_{2}\\ \mathbf{-}\mathbf{8}{\mathbf{e}}^{-2x}\mathbf{-}\mathbf{2}\left(\mathbf{4}{\mathbf{e}}^{-2x}\right)\mathbf{-}\mathbf{5}\left(\mathbf{-}\mathbf{2}{\mathbf{e}}^{-2x}\right)\mathbf{+}\mathbf{6}\left({\mathbf{e}}^{-2x}\right)\mathbf{=}\mathbf{27}{\mathbf{e}}^{3x}\mathbf{-}\mathbf{2}\left(\mathbf{9}{\mathbf{e}}^{3x}\right)\mathbf{-}\mathbf{5}\left(\mathbf{3}{\mathbf{e}}^{3x}\right)\mathbf{+}\mathbf{6}\left({\mathbf{e}}^{3x}\right)\\ \mathbf{-}\mathbf{8}{\mathbf{e}}^{-2x}\mathbf{-}\mathbf{8}{\mathbf{e}}^{-2x}\mathbf{+}\mathbf{10}{\mathbf{e}}^{-2x}\mathbf{+}\mathbf{6}{\mathbf{e}}^{-2x}\mathbf{=}\mathbf{27}{\mathbf{e}}^{3x}\mathbf{-}\mathbf{18}{\mathbf{e}}^{3x}\mathbf{-}\mathbf{15}{\mathbf{e}}^{3x}\mathbf{+}\mathbf{6}{\mathbf{e}}^{3x}\\ 0=0\end{array}$

Using the Wronskian,

$\begin{array}{c}\mathbf{w}\mathbf{=}|\begin{array}{ccc}{\mathbf{e}}^{x}& {\mathbf{e}}^{-2x}& {\mathbf{e}}^{3x}\\ {\mathbf{e}}^{x}& \mathbf{-}\mathbf{2}{\mathbf{e}}^{-2x}& \mathbf{3}{\mathbf{e}}^{3x}\\ {\mathbf{e}}^{x}& \mathbf{4}{\mathbf{e}}^{-2x}& \mathbf{9}{\mathbf{e}}^{3x}\end{array}|\\ \mathbf{=}{\mathbf{e}}^{x}\left(\mathbf{-}\mathbf{18}{\mathbf{e}}^{x}\mathbf{-}\mathbf{12}{\mathbf{e}}^{x}\right)\mathbf{-}{\mathbf{e}}^{-2x}\left(\mathbf{9}{\mathbf{e}}^{4x}\mathbf{-}\mathbf{3}{\mathbf{e}}^{4x}\right)\mathbf{+}{\mathbf{e}}^{3x}\left(\mathbf{4}{\mathbf{e}}^{-x}\mathbf{+}\mathbf{2}{\mathbf{e}}^{-x}\right)\\ \mathbf{=}\mathbf{-}\mathbf{30}{\mathbf{e}}^{2x}\\ \ne \mathbf{0}\end{array}$

Hence, ${\mathbf{w}}{\ne }{\mathbf{0}}$and now we can say that ${{\mathbf{e}}}^{x}{,}{\text{\hspace{0.17em}\hspace{0.17em}}}{{\mathbf{y}}}_{1}{,}{\text{\hspace{0.17em}\hspace{0.17em}}}{{\mathbf{y}}}_{2}$ are linearly independent solutions. ### Want to see more solutions like these? 