Question

Write y=ue^(ax) to find the general solution of y"+2y'+y=e^(-x)/(sqrt(x)).

Answer 1

\[y''+2y'+y=\frac{ e ^{-x} }{ \sqrt{x} }\]

Answer 2

@SithsAndGiggles

Answer 3

find y' and y''
plug in you should get something like:
\[e^{ax}[ u''+2u'(a+1)+u(a+1)^2]=e^{-x} x^{\frac{-1}{2}} \\ \text{ compare } e^{ax} \text{ to } e^{-x} \text{ so } a=-1 \\ \text{ plug in } -1 \text{ for } a \\ u''=x^{\frac{-1}{2}}\]

Answer 4

let me know if you need further help or if you don't get any of that?

Answer 5

Thank you!