Question

Answer Problems 1-10 without referring back to the text. Fill in the blank or answer true or false.

$y_p=A{x}^2$is particular solution of $y^{¢¢¢}+y^{¢¢}=1$for A=

Short answer

A=$\frac{1}{2}$

Step-by-step solution

Definition

A Particular Solution of a differential equation is a solution obtained from the General Solution by assigning specific values to the arbitrary constants.

Find value of A

Consider the differential equation$y^{m\text{'}\text{'}\text{'}}+y^{\text{'}}=1..........\left(1\right)$

We must find the derivatives of $y_p=A{x}^2$and plug them into equation (1) in order to solve for A

localid="1668502520667" $$\begin{gathered} y^{\text{'}}=2Ax \\ {y}^{\text{'}\text{'}}=2A \\ {y}^{\text{'}\text{'}\text{'}}=0 \end{gathered}$$

Substituting values in equation (1), we have

0+2A=1

A =$\frac{1}{2}$