Question

Find a linear second-order differential equation with constant coefficients for which ${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{1}$ and ${\mathit{y}}_{\mathbf{2}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$are solutions of the associated homogeneous equation and ${\mathit{y}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathit{x}$is a particular solution of the non homogeneous equation.

Short answer

A linear second-order differential equation$y^{\text{'}\text{'}}-y^{\text{'}}=\frac{1}{2}{x}^2-x$

Step-by-step solution

Define the particular solution 

A Particular Solution is a differential equation solution that is derived from the General Solution by assigning specific values to the random constants. Depending on the query, the requirements for finding the values of the random constants can be submitted to us as an Initial-Value Problem or Boundary Conditions.

Now find the second order differential equation when we have the particular solution yp=12x2-x

Assume that$y=e^{rx}$ is a solution of the required differential equation.

Second,

role="math" localid="1667905807698" $$\begin{gathered} y_1=1=e^0=e^{r_{1}x} \\ {y}_2=e^x=e^{r_{2}x} \end{gathered}$$

Then we have the roots

$r_1=0\text{and}{m}_2=1$

Then we can have

role="math" localid="1667905883759" $$\begin{gathered} r(r-1)=0 \\ {r}^2-r=0 \end{gathered}$$

The auxiliary solution for the required differential equation.

Then we can obtain the corresponding homogeneous equation as

$y^{\text{'}\text{'}}-y^{\text{'}}=0$

Finally, since we have $y_p=\frac{1}{2}{x}^2-x,$

The non-homogeneous differential equation as

$y^{\text{'}\text{'}}-y^{\text{'}}=\frac{1}{2}{x}^2-x$