Question

A differential equation may possess more than one family of solutions.

(a) Plot different members of the families$\mathit{y}\mathbf{=}{\mathit{\varphi }}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{c}}_{\mathbf{1}}$ and$\mathit{y}\mathbf{=}{\mathit{\varphi }}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}$.

(b) Verify that$\mathit{y}\mathbf{=}{\mathit{\varphi }}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$and$\mathit{y}\mathbf{=}{\mathit{\varphi }}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$are two solutions of the nonlinear first-order differential equation$\mathbf{(}\mathbf{y}\mathbf{\text{'}}{\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{4}{\mathit{x}}^{\mathbf{2}}$.

(c) Construct a piecewise-defined function that is a solution of the nonlinear DE in part (b) but is not a member of either family of solutions in part (a).

Short answer

(a) The graph is shown below.

(b) Both are two solutions of the nonlinear first-order DE.

(c) $y\left(\mathrm{x}\right)=\{\begin{array}{l}-{\mathrm{x}}^2\mathrm{x}\le 0\\ {\mathrm{x}}^2\mathrm{x}\ge 0\end{array}$is a solution of given DE.

Step-by-step solution

Define a derivative of the function

The sensitivity of the function is derived by the derivative of a function of a real variable for modifying the argument.

Calculus uses derivatives as a fundamental tool. When a derivative of a single-variable function exists at a given input value, it is the slope of the tangent line to the function's graph at that point.

Step 2(a): The graph of the different members of the family

Let the graph of the different members of the family $\mathrm{y}={\varphi }_1\left(\mathrm{x}\right)$and $\mathrm{y}={\varphi }_2\left(\mathrm{x}\right)$be,

Hence the graph is drawn.

Step 3(b): Verification of the solutions

As for the first solution,$\mathrm{y}={\varphi }_1\left(\mathrm{x}\right)={\mathrm{x}}^2+{\mathrm{c}}_1$, substitute$\mathrm{y}\text{'}=2\mathrm{x}$in it to get,

$(y\text{'}{)}^2=(2x{)}^2=4x^2$

As for the second solution, $\mathrm{y}={\varphi }_2\left(\mathrm{x}\right)=-{\mathrm{x}}^2+{\mathrm{c}}_2$, substitute $\mathrm{y}\text{'}=-2\mathrm{x}$in it to get,

$(y\text{'}{)}^2=(-2x{)}^2=4x^2$

Hence, this shows that both are solutions of the first-order DE.

Step 4(c): Construct a piecewise-defined function

As for the nonlinear DE, using the two solutions together, the required piecewise-defined function is given by,

$y\left(\mathrm{x}\right)=\{\begin{array}{l}-{\mathrm{x}}^2\mathrm{x}\le 0\\ {\mathrm{x}}^2\mathrm{x}\ge 0\end{array}$

For the above equation,

$y\text{'}\left(\mathrm{x}\right)=\{\begin{array}{l}-2\mathrm{x}\mathrm{x}\le 0\\ 2\mathrm{x}\mathrm{x}\ge 0\end{array}$

And,

$y\text{'}\text{'}\left(\mathrm{x}\right)=\{\begin{array}{l}-2\mathrm{x}\le 0\\ 2\mathrm{x}\ge 0\end{array}$

Hence, $y\left(\mathrm{x}\right)=\{\begin{array}{l}-{\mathrm{x}}^2\mathrm{x}\le 0\\ {\mathrm{x}}^2\mathrm{x}\ge 0\end{array}$ is a solution to given DE.