Question

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$\mathbf{w}\mathbf{=}\mathbf{v}\mathbf{\text{'}}$.

Short answer

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

Step-by-step solution

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$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}{\mathbf{x}}^2\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{=}\mathbf{0}\mathrm{......}\left(\mathbf{1}\right)$

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,

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

Conclusion

Substitute the all values in the equation (1),

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

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