Question

Verify the statement of Example 2. Also verify that $\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{}\mathit{x}$and$\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{}\mathit{x}$ are solutions of ${\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{=}\mathit{y}$.

Short answer

It is verified that $\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{}\mathit{x}$and $\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{}\mathit{x}$are the solutions for the differential equation ${\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{=}\mathit{y}$.

Step-by-step solution

Given information

The given differential equation is${\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{=}\mathit{y}$.

Meaning of differential equation

In mathematics, an equation with only one independent variable and one or more of its derivatives with respect to the variable is referred to as anordinarydifferentialequation or ODE.In other words, the ODE is a relation with one independent variable x, a real dependent variable y, and several derivatives ${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{,}\mathbf{.}\mathbf{...}\mathbf{,}{\mathit{y}}^{\mathbf{n}}$in relation to x.

Verify that y=sinhxand are 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Example 2 states that the differential equation $y^{\text{'}\text{'}}=y$has solution $y=e^x$,$y=e^{-x}$, or $y=A{e}^x+B{e}^{-x}$.Verify this statement by substituting these solutions into the differential equation.

Let’s start with $y=e^x$.

$\begin{array}{l}{y}^{\text{'}}=e^x\\ y^{\text{'}\text{'}}=e^x\end{array}$

Substitutethisin the differential equation.

$e^x=e^x$

For,$y=e^{-x}$,

$\begin{array}{l}{y}^{\text{'}}=-e^{-x}\\ y^{\text{'}\text{'}}=e^{-x}\end{array}$

Substitutethisin the differential equation.

$e^{-x}=e^{-x}$

For $y=A{e}^{x}B{e}^{-x}$,

$\begin{array}{l}{y}^{\text{'}}=A{e}^x-B{e}^{-x}\\ y^{\text{'}\text{'}}=A{e}^x+B{e}^{-x}\end{array}$.

.Again substitute this in the differential equation.

$A{e}^x+B{e}^{-x}=A{e}^x+B{e}^{-x}$

Now, verify that there are two solutions for the same differential equation.

The first one is .

localid="1654779330725" $\begin{array}{l}{y}^{\text{'}}=\sinh x\\ y^{\text{'}\text{'}}=\cosh x\end{array}$

Substitute this into the given differential equation.

localid="1654779345044" $\cosh x=\cosh x$

The second one is.localid="1654779359740" $y=\sinh x$

$\begin{array}{l}{y}^{\text{'}}=\cosh x\\ y^{\text{'}\text{'}}=\sinh x\end{array}$

Substitute this into the given differential equation.

localid="1654779380803" $\sinh x=\sinh x$

Hence, proved.