Question

$y_p=A{x}^2$is particular solution of$y\text{'}\text{'}\text{'}+y\text{'}\text{'}=1$for$A=\_\_\_\_$

Short answer

The value of A is $A=\frac{1}{2}$.

Step-by-step solution

Define a particular solution of the differential equation.

A solution of the form $y=f\left(x\right)$is a differential equation solution that does not contain any arbitrary constants. The differential equation's general solution is of the form$y=f\left(x\right)$ or$y=ax+b$ , with and being arbitrary constants.

A solution of the differential equation is y=-5e-x+10ex.

Let the particular solutions be, ${\text{y}}_{\text{p}}{\text{= Ax}}^{\text{2}}$.

Differentiate with respect to x .

$$\begin{gathered} {\text{y'}}_p=2Ax \\ {\text{y}}_p^{\text{'}\text{'}}=2A \\ {\text{y}}_p^{\text{'}\text{'}\text{'}}=0 \end{gathered}$$

Substitute the values in the equation $\text{y''' + y'' = 1}$.

$$\begin{gathered} \text{0 + 2A = 1} \\ \text{2A = 1} \\ \text{A =}\frac{1}{2} \end{gathered}$$