Question

variation of parameter

(D^2 - 1)y = e^x + 1

Answer 1

\[\large{y''-y=e^x+1}\]first find the homogeneuos solution\[\large{y''-y}=0\]\[\large{r^2-1}=0\]\[\large{(r-1)(r+1)=0}\]\[\large{r=1,-1}\]\[\large{y_h=c_1e^x+c_2e^{-x}}\]now to find the particular solution using variation of parameters
first find wronskian\[\large{y_p=-y_1 \int\limits{\frac{y_2(e^x+1)}{W}dx}+y_2 \int\limits{\frac{y_1(e^x+1)}{W}dx}}\]where y1=e^x and y2=e^-x