Question

Given that $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}$ is a two-parameter family of solutions of$\mathbf{xy}\mathbf{\text{'}\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$ on the interval$\left(-\infty ,\infty \right)$, show that constants${\mathbf{c}}_{\mathbf{1}}$and${\mathbf{c}}_{\mathbf{2}}$cannot be found so that a member of the family satisfies the initial conditions$\mathbf{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$. Explain why this does not violate Theorem 4.1.1.

Short answer

The given equation does not satisfies the Theorem 4.1.1

Step-by-step solution

Theorem 4.1.1

${\mathrm{a}}_{\mathrm{n}}\left(\mathrm{x}\right),{\mathrm{a}}_{\mathrm{n}-1}\left(\mathrm{x}\right),...,{\mathrm{a}}_1\left(\mathrm{x}\right),{\mathrm{a}}_0\left(\mathrm{x}\right)$and $\mathrm{g}\left(\mathrm{x}\right)$I and let be continuous at intervals. For every x in this interval ${\mathrm{a}}_{\mathrm{n}}\left(\mathrm{x}\right)\ne 0$. If $\mathrm{x}={\mathrm{x}}_0$is any point in this interval, then the solution to the initial-value problem (1) lies in the$\mathrm{y}\left(\mathrm{x}\right)$ interval and is unique.

Prove the given equation satisfies the condition

$\mathrm{y}={\mathrm{c}}_1+{\mathrm{c}}_2{\mathrm{x}}^2$

Consider $\mathrm{y}\left(0\right)=0$,

${\mathrm{c}}_1+{\mathrm{c}}_2{\left(0\right)}^2=0$

Let${\mathrm{c}}_1=0\mathrm{and}\mathrm{y}\text{'}\left(0\right)=1$

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{x}\right)=2{\mathrm{c}}_2\mathrm{x} \\ \mathrm{y}\text{'}\left(0\right)=2{\mathrm{c}}_2\left(0\right) \\ 1=0 \end{gathered}$$

Where 1≠0,

Then ${\mathrm{c}}_1$and ${\mathrm{c}}_2$cannot be determined. So the member of the family satisfies the initial conditions.

To prove the theorem 4.1.1,

Consider the general second order differential equation,

${\mathrm{a}}_2\left(\mathrm{x}\right)\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+{\mathrm{a}}_1\left(\mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}+{\mathrm{a}}_0\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)$

Comparing with IVP we get,

$\mathrm{x}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\frac{\mathrm{dy}}{\mathrm{dx}}=0$

By comparing we get,

${\mathrm{a}}_2\left(\mathrm{x}\right)=\mathrm{x},{\mathrm{a}}_1\left(\mathrm{x}\right)=-1,{\mathrm{a}}_0\left(\mathrm{x}\right)=0,\mathrm{g}\left(\mathrm{x}\right)=0.$

According to the Theorem 4.1.1 given in step 1,

${\mathrm{a}}_2\left(\mathrm{x}\right)\ne 0$but we have got the solution as ${\mathrm{a}}_2\left(\mathrm{x}\right)=\mathrm{x}$is equal to 0 for $\mathrm{x}=0$

Hence the equation does not fulfill the condition of theorem 4.1.1.