Question

In Problems 13 and 14 the given two-parameter family is a solution of the indicated differential equation on the interval $\left(-\infty ,\infty \right)$. Determine whether a member of the family can be found that satisfies the boundary conditions.

role="math" localid="1667889531022" $$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{3}\mathbf{;}\mathbf{}{\mathbf{x}}^{\mathbf{2}}\mathbf{}\mathbf{y}\mathbf{\text{'}\text{'}}\mathbf{-}\mathbf{5}\mathbf{x}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{=}\mathbf{24} \\ \mathbf{}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{y}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{4} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{2} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{=}\mathbf{15} \end{gathered}$$

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{No}\mathrm{solution} \\ \left(\mathrm{b}\right)\mathrm{No}\mathrm{solution} \\ \left(\mathrm{c}\right)\mathrm{y}={\mathrm{c}}_1{\mathrm{x}}^2-\left({\mathrm{c}}_1+3\right){\mathrm{x}}^4+3 \\ \left(\mathrm{d}\right)\mathrm{y}=-{\mathrm{x}}^2+{\mathrm{x}}^4+3 \end{gathered}$$

Step-by-step solution

Step 1(a): Determine whether a member of the family can be found that satisfies the boundary condition y(-1)=0, y(1)=4

$$\begin{gathered} \mathrm{y}\left(-1\right)=0\mathrm{and}\mathrm{y}\left(1\right)=4\mathrm{gives} \\ ={\mathrm{c}}_1{\left(-1\right)}^2+{\mathrm{c}}_2{\left(-1\right)}^4+3 \\ ={\mathrm{c}}_1·1+{\mathrm{c}}_2·1+3 \\ ={\mathrm{c}}_1+{\mathrm{c}}_2+3 \\ \mathrm{y}\left(1\right)=4 \\ ={\mathrm{c}}_1{\left(1\right)}^2+{\mathrm{c}}_2{\left(1\right)}^4+3 \\ ={\mathrm{c}}_1·1+{\mathrm{c}}_2·1+3 \\ ={\mathrm{c}}_1+{\mathrm{c}}_2+3 \\ \mathrm{Since}0\ne 4 \end{gathered}$$

Hence, there is no member of the family using these initial conditions.

Step 2(b): Determine whether a member of the family can be found that satisfies the boundary condition y(0)=1, y(1)=2

$$\begin{gathered} \mathrm{y}\left(0\right)=1\mathrm{and}\mathrm{y}\left(1\right)=2\mathrm{gives} \\ \mathrm{y}\left(0\right)={\mathrm{c}}_1{\left(0\right)}^2+{\mathrm{c}}_2{\left(0\right)}^4+3 \\ 1={\mathrm{c}}_1·0+{\mathrm{c}}_2·0+3 \\ 1=3 \end{gathered}$$

which is not true. This implies that there is no member of the family using these initial conditions.

Hence, there is no member of the family using these initial conditions.

Step 3(c): Determine whether a member of the family can be found that satisfies the boundary condition y(0)=3, y(1)=0

$$\begin{gathered} \mathrm{y}\left(0\right)=3\mathrm{and}\mathrm{y}\left(1\right)=0\mathrm{gives} \\ \mathrm{y}\left(0\right)={\mathrm{c}}_1\left(0^2\right)+{\mathrm{c}}_2{\left(0\right)}^4+3 \\ 3={\mathrm{c}}_1·0+{\mathrm{c}}_2·0+3 \\ 3=3 \\ \mathrm{y}\left(1\right)={\mathrm{c}}_1{\left(1\right)}^2+{\mathrm{c}}_2{\left(1\right)}^4+3 \\ 0={\mathrm{c}}_1·1+{\mathrm{c}}_2·1+3 \\ 0={\mathrm{c}}_1+{\mathrm{c}}_2+3 \\ {\mathrm{c}}_2=-\left({\mathrm{c}}_1+3\right) \end{gathered}$$

Where $c_1$can be any constant.

Therefore, the solution is$\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\left(c_1+3\right){\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{3}$

Step 4(d): Determine whether a member of the family can be found that satisfies the boundary condition y(1)=3, y(2)=15

$$\begin{gathered} \mathrm{y}\left(1\right)=3\mathrm{and}\mathrm{y}\left(2\right)=15\mathrm{gives} \\ \mathrm{y}\left(1\right)={\mathrm{c}}_1{\left(1\right)}^2+{\mathrm{c}}_2{\left(1\right)}^4+3 \\ 3={\mathrm{c}}_1·1+{\mathrm{c}}_2·1+3 \\ 3={\mathrm{c}}_1+{\mathrm{c}}_2+3 \\ 0={\mathrm{c}}_1+{\mathrm{c}}_2 \\ {\mathrm{c}}_2=-{\mathrm{c}}_1 \\ \mathrm{y}\left(2\right)={\mathrm{c}}_1{\left(2\right)}^2+{\mathrm{c}}_2{\left(2\right)}^4+3 \\ 15={\mathrm{c}}_1·4-{\mathrm{c}}_1·16+3\mathrm{Substituting}{\mathrm{c}}_2=-{\mathrm{c}}_1 \\ 15=4{\mathrm{c}}_1-16{\mathrm{c}}_1+3 \\ 12=-12{\mathrm{c}}_1 \\ {\mathrm{c}}_1=-1 \\ \mathrm{This}\mathrm{implies}{\mathrm{c}}_2=-{\mathrm{c}}_1=1. \end{gathered}$$

Therefore, the solution is$\mathbf{y}\mathbf{=}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{3}$