Question

In Problems 1 and 2 verify that and are solutions of the given differential equation but that is, in general, not a solution.

1.
${\left(y^{\text{'}\text{'}}\right)}^{\mathbf{2}}\mathbf{=}{\mathit{y}}^{\mathbf{2}}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{1}}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{,}{\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{x}$

Short answer

The solution is

Verify by substituting values of y₁ and y₂ the second derivative of y.

Step-by-step solution

given information

${\left(y^{\text{'}\text{'}}\right)}^2=y^2;y_1=e^x,y_2=cosx$

solve second order differential equation

We have the second order differential equation

${\left(y^{\text{'}\text{'}}\right)}^2=y^2.....\left(1\right)$

With the alleged solutions $y_1=e^{x}and{y}_2=cosx$ we have to verify that they are solutions or not as the following

For the alleged solutions $y_1=e^x$ differentiate with respect to x, then we have

$$\begin{gathered} y_1^{\text{'}}=e^x \\ {y}_1^{\text{'}\text{'}}=e^x...\left(2\right) \end{gathered}$$

After that, substitute with $y_1^{\text{'}\text{'}}$ shown in equation (2) into equation (1), then we have

$$\begin{gathered} {\left(e^x\right)}^2=e^{2x} \\ =y_1^2 \end{gathered}$$

Then $y_1=e^x$is a solution of the given differential equation

For the second alleged solution $y_2=cosx$: differentiate with respect to x , then we have

$$\begin{gathered} y_2^{\text{'}}=-sinx \\ {y}_2^{\text{'}\text{'}}=-cosx...\left(3\right) \end{gathered}$$

After that, substitute with y'' shown in equation (3) into equation (1), then we have

$$\begin{gathered} (-cosx{)}^2=co{s}^{2}x \\ =(cosx{)}^2 \\ =y_2^2 \end{gathered}$$

Then $y_2=cosx$is a solution of the given differential equation.

verify

Now if we considered the total solution $y=c_1{e}^x+c_{2}cosx$, we have to verify that it is not a solution in general as the following:

Differentiate with respect to x, then we have

$$\begin{gathered} y\text{'}=c_1{e}^x-c_{2}sinx{y}^{\text{'}\text{'}} \\ \text{y''}=c_1{e}^x-c_{2}cosx....\left(4\right) \end{gathered}$$

After that, substitute with and shown in equation (4) into equation (1), then we have

$$\begin{gathered} {\left(c_1{e}^x-c_{2}cosx\right)}^2\ne {\left(c_1{e}^x+c_{2}cosx\right)}^2 \\ \ne y^2 \end{gathered}$$

then $\left(y=c_1{e}^x+c_{2}cosx\right)$is not a solution of the given differential equation.

conclusion

Verify by substituting values of y₁ and y₂ the second derivative of y.