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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

# Given that the function ${\mathbf{f}}\left(\mathbf{x}\right){\mathbf{=}}{\mathbf{x}}$ is a solution to ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{{\mathbf{x}}}^{2}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{xy}}{\mathbf{=}}{\mathbf{0}}$, show that the substitution ${\mathbf{y}}\left(\mathbf{x}\right){\mathbf{=}}{\mathbf{v}}\left(\mathbf{x}\right){\mathbf{f}}\left(\mathbf{x}\right){\mathbf{=}}{\mathbf{v}}\left(\mathbf{x}\right){\mathbf{x}}$ reduces this equation to, ${\mathbf{xw}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{3}}{\mathbf{w}}{\mathbf{\text{'}}}{\mathbf{-}}{{\mathbf{x}}}^{3}{\mathbf{w}}{\mathbf{=}}{\mathbf{0}}$ where ${\mathbf{w}}{\mathbf{=}}{\mathbf{v}}{\mathbf{\text{'}}}$.

Thus, it is proved that the given equation can be reduced to$\mathbf{xw}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{w}\mathbf{\text{'}}\mathbf{-}{\mathbf{x}}^{3}\mathbf{w}\mathbf{=}\mathbf{0}.$

See the step by step solution

## Step 1: Use the given functions to reduce the given equation to xw''+3w'-x3w=0

Given that $\mathbf{f}\left(\mathbf{x}\right)\mathbf{=}\mathbf{x}$ is a solution to $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}{\mathbf{x}}^{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{=}\mathbf{0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(\mathbf{1}\right)$

And

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

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

$\begin{array}{c}y=\mathrm{vx}\\ y\text{'}=v+\mathrm{xv}\text{'}\end{array}$

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

$\begin{array}{c}y\text{'}=v+\mathrm{xw}\\ y\text{'}\text{'}=v\text{'}+\mathrm{xw}\text{'}+w\\ y\text{'}\text{'}=w+\mathrm{xw}\text{'}+w\\ y\text{'}\text{'}=2w+\mathrm{xw}\text{'}\\ y\text{'}\text{'}\text{'}=2w\text{'}+w\text{'}+\mathrm{xw}\text{'}\text{'}\\ y\text{'}\text{'}\text{'}=3w\text{'}+\mathrm{xw}\text{'}\text{'}\end{array}$

## Step 2: Conclusion

Substitute the all values in the equation (1),

$\begin{array}{c}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}{\mathbf{x}}^{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{=}\mathbf{0}\\ \mathbf{3}\mathbf{w}\mathbf{\text{'}}\mathbf{+}\mathbf{xw}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}{\mathbf{x}}^{2}\left(\mathbf{v}\mathbf{+}\mathbf{xw}\right)\mathbf{+}\mathbf{x}\left(\mathbf{vx}\right)\mathbf{=}\mathbf{0}\\ \mathbf{3}\mathbf{w}\mathbf{\text{'}}\mathbf{+}\mathbf{xw}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}{\mathbf{x}}^{2}\mathbf{v}\mathbf{-}{\mathbf{x}}^{3}\mathbf{w}\mathbf{+}{\mathbf{x}}^{2}\mathbf{v}\mathbf{=}\mathbf{0}\\ \mathbf{3}\mathbf{w}\mathbf{\text{'}}\mathbf{+}\mathbf{xw}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}{\mathbf{x}}^{3}\mathbf{w}\mathbf{=}\mathbf{0}\\ \mathbf{xw}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{w}\mathbf{\text{'}}\mathbf{-}{\mathbf{x}}^{3}\mathbf{w}\mathbf{=}\mathbf{0}\end{array}$

Thus, it is proved that the given equation can be reduced to $\mathbf{xw}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{w}\mathbf{\text{'}}\mathbf{-}{\mathbf{x}}^{3}\mathbf{w}\mathbf{=}\mathbf{0}.$