Question

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

2.

$\mathit{y}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\left(y^{\text{'}}\right)}^{\mathbf{2}}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{y}}_{\mathbf{2}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}$

Short answer

The solution is

Verify by substituting values of y₁and y₂the second derivative of y.

Step-by-step solution

given information

$y{y}^{\text{'}\text{'}}=\frac{1}{2}{\left(y^{\text{'}}\right)}^2;y_1=1,y_2=x^2$

solve second order differential equation

We have the second order differential equation

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

With the alleged solutions role="math" localid="1668074050949" $y_1=1and{y}_2=x^2$, we have to verify that they are solutions or not as the following

For the alleged solutions y₁ = 1 differentiate with respect to x, then we have

$$\begin{gathered} y_1^{\text{'}}=0....\left(2\right) \\ {y}_1^{\text{'}\text{'}}=0....\left(3\right) \end{gathered}$$

After that, substitute with and shown in equation (2) into equation (1), then we have

$$\begin{gathered} 1\times 0=0 \\ =\frac{1}{2}{\left(y^{\text{'}}\right)}^2 \end{gathered}$$

Then is a solution of the given differential equation

For the second alleged solution $y_2=x^2$: differentiate with respect to x, then we have

$$\begin{gathered} y_2^{\text{'}}=2x....\left(4\right) \\ {y}_2^{\text{'}\text{'}}=2....\left(5\right) \end{gathered}$$

After that, substitute with and shown in equations (4) and (5) into equation (1), then we have

$$\begin{gathered} x^2\times 2=2x^2 \\ =\frac{1}{2}\left(4x^2\right) \\ =\frac{1}{2}(2x{)}^2 \\ =\frac{1}{2}{\left(y^{\text{'}}\right)}^2 \end{gathered}$$

Then $y_2=x^2$is a solution of the given differential equation.

verify

Now if we considered the total solution $y=c_1+c_2{x}^2$, we have to verify that it is not a solution in general as the following:

Differentiate with respect to x, then we have

$$\begin{gathered} y^{\text{'}}=2c_{2}x....\left(6\right) \\ {y}^{\text{'}\text{'}}=2c_2......\left(7\right) \end{gathered}$$

After that, substitute with and shown in equation (6) and (7) into equation (1) , then we have

$$\begin{gathered} \left(c_1+c_2{x}^2\right)\left(2c_2\right)=2c_1{c}_2+2c_2^2{x}^2 \\ \ne \frac{1}{2}{\left(2c_{2}x\right)}^{2}n \\ \ne \frac{1}{2}{\left(y^{\text{'}}\right)}^2 \end{gathered}$$

then $y=c_1+c_2{x}^2$is not a solution of the given differential equation.

conclusion

Verify by substituting values of y₁and y₂the second derivative of y.