Question

Use the result of Problem 34 to construct a third-order differential equation for which$\{\mathbf{x}\mathbf{,}\mathbf{sinx}\mathbf{,}\mathbf{cosx}\}$ is a fundamental solution set.

Short answer

Thus, the required differential equation is$\mathbf{xy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}.$

Step-by-step solution

Use the given fundamental solution set,

Given the fundamental solution set,

$\{\mathbf{x}\mathbf{,}\mathbf{sinx}\mathbf{,}\mathbf{cosx}\}$

We have,

${\mathbf{f}}_1\mathbf{=}\mathbf{x}\mathbf{,}{\mathbf{f}}_2\mathbf{=}\mathbf{sinx}\mathbf{,}{\mathbf{f}}_3\mathbf{=}\mathbf{cosx}$

Given the equation of problem 34 is,

$\left|\begin{array}{cccc}{\mathbf{f}}_1\left(\mathbf{x}\right)& {\mathbf{f}}_2\left(\mathbf{x}\right)& {\mathbf{f}}_3\left(\mathbf{x}\right)& y\\ {\mathbf{f}}_1\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\left(\mathbf{x}\right)& y\text{'}\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& y\text{'}\text{'}\\ {\mathbf{f}}_1\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_2\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& {\mathbf{f}}_3\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)& y\text{'}\text{'}\text{'}\end{array}\right|\mathbf{=}\mathbf{0}\mathrm{.......}\left(\mathbf{1}\right)$

Now Use the given equation of Problem 34,

Substitute ${\mathbf{f}}_1\mathbf{=}\mathbf{x}\mathbf{,}{\mathbf{f}}_2\mathbf{=}\mathbf{sinx}\mathbf{,}{\mathbf{f}}_3\mathbf{=}\mathbf{cosx}$in the equation (1),

$\left|\begin{array}{cccc}x& \mathrm{sinx}& \mathrm{cosx}& y\\ 1& \mathrm{cosx}& -\mathrm{sinx}& y\text{'}\\ 0& -\mathrm{sinx}& -\mathrm{cosx}& y\text{'}\text{'}\\ 0& -\mathrm{cosx}& \mathrm{sinx}& y\text{'}\text{'}\text{'}\end{array}\right|\mathbf{=}\mathbf{0}$

Now find the determinate in the above equation,

$\begin{array}{c}\mathbf{x}\left|\begin{array}{ccc}\mathrm{cosx}& -\mathrm{sinx}& y\text{'}\\ -\mathrm{sinx}& -\mathrm{cosx}& y\text{'}\text{'}\\ -\mathrm{cosx}& \mathrm{sinx}& y\text{'}\text{'}\text{'}\end{array}\right|\mathbf{-}\mathbf{1}\left|\begin{array}{ccc}\mathrm{sinx}& \mathrm{cosx}& y\\ -\mathrm{sinx}& -\mathrm{cosx}& y\text{'}\text{'}\\ -\mathrm{cosx}& \mathrm{sinx}& y\text{'}\text{'}\text{'}\end{array}\right|\mathbf{=}\mathbf{0}\\ \mathbf{x}(\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}})\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\\ -\mathrm{xy}\text{'}\text{'}\text{'}-\mathrm{xy}\text{'}+y\text{'}\text{'}+y=0\\ \mathrm{xy}\text{'}\text{'}\text{'}+\mathrm{xy}\text{'}-y\text{'}\text{'}-y=0\end{array}$

Thus, the required differential equation is$\mathbf{xy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}.$