Given that $y_1,y_2$ be a fundamental set of solutions of the ODE and $(\mathrm{e}^xy')'+y'+y=0, x>0$ if the Wronskian $W(y_1,y_2)(0)=1$ then $W(y_1,y_2)(-1)=? $
my attempt:
Given D.e can be written as $\mathrm{e}^xy''+(\mathrm{e}^x+1)y'+y=0$
hence by Abel's formula $W(y_1,y_2)(x)=c\,\mathrm{e}^{-\int \frac{\mathrm{e}^x+1}{\mathrm{e}^x}}dx=c\,\mathrm{e}^{-\frac{(\mathrm{e}^x+1)^2}{2}}$
Since $W(y_1,y_2)(0)=1 \implies c=\mathrm{e}^2$ , I getting doubt I am missing something can some one help please
Best Answer
$W(y_1,y_2)(x)=ce^{-\int \frac{e^x+1}{e^x} dx}=ce^{-\int 1+ \frac{1}{e^x} dx}=c e^{(e^{-x}-x)}$
$W(y_1,y_2)(0)=1 \implies c e^{(1-0)}=1 \implies c=\frac{1}{e}$
$\therefore W(y_1,y_2)(-1)=\frac{1}{e} e^{(e^{1}+1)}=e^{(e+1-1)}=e^e$