Verify that the function $\mathbf{\varphi }\left(x\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}$is a solution to the linear equation $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$ for any choice of the constants ${\mathbf{c}}_{\mathbf{1}}$ and${\mathbf{c}}_{\mathbf{2}}$. Determine ${\mathbf{c}}_{\mathbf{1}}$and ${\mathbf{c}}_{\mathbf{2}}$so that each of the following initial conditions is satisfied.

(a)$\mathbf{y}\left(0\right)\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$

(b)$\mathbf{y}\left(1\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}}\left(1\right)\mathbf{=}\mathbf{0}$

First of all, take the given function as,$\varphi \left(\mathrm{x}\right)=\mathrm{y}$.

Differentiating $\varphi \left(\mathrm{x}\right)=\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}$, concerning x,

$\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}$

Again, differentiating concerning x,

$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+4{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}$

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+4{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}} \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+4{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}-{\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+2{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}} \\ \begin{array}{l}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+4{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}-\left({\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}\right)+{\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}\\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+4{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}-\frac{\mathrm{dy}}{\mathrm{dx}}+{\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}\\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=2{\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+2{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}-\frac{\mathrm{dy}}{\mathrm{dx}}\\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=2\left({\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-2\mathrm{x}}\right)-\frac{\mathrm{dy}}{\mathrm{dx}}\\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=2\mathrm{y}-\frac{\mathrm{dy}}{\mathrm{dx}}\\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=2\mathrm{y}-\frac{\mathrm{dy}}{\mathrm{dx}}\\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+\frac{\mathrm{dy}}{\mathrm{dx}}-2\mathrm{y}=0\end{array} \end{gathered}$$

Which is identical to the given differential equation.

Hence, $\varphi \left(\mathbf{x}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}$ is a solution to $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$, for any choice of the constants ${\mathbf{c}}_1$ and ${\mathbf{c}}_2$.

As given in part (a) of the question that when x = 0, y = 2

Therefore, we will put these values in the given function $\mathrm{\varphi }\left(\mathrm{x}\right)$,

$\begin{array}{l}{\mathrm{c}}_1{\mathrm{e}}^0+{\mathrm{c}}_2{\mathrm{e}}^0=2\\ {\mathrm{c}}_1+{\mathrm{c}}_2=2······\left(1\right)\end{array}$

Also, when$\mathrm{x}=0,\mathrm{y}\text{'}=1$

Therefore, we will put these values in $\frac{\mathrm{dy}}{\mathrm{dx}}$,

$\begin{array}{l}{\mathrm{c}}_1{\mathrm{e}}^0-2{\mathrm{c}}_2{\mathrm{e}}^0=1\\ {\mathrm{c}}_1-2{\mathrm{c}}_2=1······\left(2\right)\end{array}$

Now, Subtracting (2) from (1),

$\begin{array}{l}3{\mathrm{c}}_2=1\\ {\mathrm{c}}_2=\frac{1}{3}\end{array}$

Putting this value of ${\mathrm{c}}_2$in (1),

$\begin{array}{l}{\mathrm{c}}_1+\frac{1}{3}=2\\ {\mathrm{c}}_1=2-\frac{1}{3}\\ {\mathrm{c}}_1=\frac{6-1}{3}\\ {\mathrm{c}}_1=\frac{5}{3}\end{array}$

Thus, to satisfy the initial condition given in part (a), ${\mathbf{c}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{5}}{\mathbf{3}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}$.

As given in part (b) of the question that when x = 1, y = 1

So, we will put these values in the given function $\mathrm{\varphi }\left(\mathrm{x}\right)$,

${\mathrm{c}}_1\mathrm{e}+{\mathrm{c}}_2{\mathrm{e}}^{-2}=1······\left(3\right)$

Also, when $\mathrm{x}=1,\mathrm{y}\text{'}=0$

Consequently, we will put these values in $\frac{\mathrm{dy}}{\mathrm{dx}}$,

${\mathrm{c}}_1\mathrm{e}-2{\mathrm{c}}_2{\mathrm{e}}^{-2}=0······\left(4\right)$

Now, Subtracting (4) from (3),

$$\begin{gathered} 3{\mathrm{c}}_2{\mathrm{e}}^{-2}=1 \\ {\mathrm{c}}_2=\frac{1}{3{\mathrm{e}}^{-2}} \end{gathered}$$

Putting this value of ${\mathrm{c}}_2$in (3),

$\begin{array}{l}{\mathrm{c}}_1\mathrm{e}+\frac{1}{3{\mathrm{e}}^{-2}}{\mathrm{e}}^{-2}=1\\ {\mathrm{c}}_1\mathrm{e}+\frac{1}{3}=1\\ {\mathrm{c}}_1\mathrm{e}=1-\frac{1}{3}\\ {\mathrm{c}}_1\mathrm{e}=\frac{2}{3}\\ {\mathrm{c}}_1=\frac{2}{3\mathrm{e}}\end{array}$

Accordingly, to satisfy the initial condition given in part (b),${\mathbf{c}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}\mathbf{e}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{2}}}$.