−y=x IF=e∫−1.dx=e−x Hence, the solution of the given differential equation is ye−x=∫xe−xdx+C ye−x=−xe−x−∫1.−e−xdx+C ye−x=−xe−x−e−x+C y=−x−1+Cex ‘y’ is not in g(x)+c form. Therefore, we can observe that statement 1 is false. Consider statement 2. (
dy
dx
)2=y order = 1,degree = 2 Therefore,statement 2 is correct