$\displaystyle y''+(\frac{x}{1+x})y'+(\frac{1}{1+x})y=0$

for x>-1

i dont know how to solve it

i was told to find a solution by the form of y1=e^x

and getting a solution

how?

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- Dec 8th 2010, 09:59 AMtransgalacticfractured mixed equations 3 c
$\displaystyle y''+(\frac{x}{1+x})y'+(\frac{1}{1+x})y=0$

for x>-1

i dont know how to solve it

i was told to find a solution by the form of y1=e^x

and getting a solution

how? - Dec 10th 2010, 03:49 PMJester
If you multily by $\displaystyle 1 + x$ then

$\displaystyle (1+x)y'' + xy' + y = 0$. Now $\displaystyle \dfrac{d}{dx}(xy'-y) = x y'' $ and $\displaystyle \dfrac{d}{dx} (xy) = x y' + y$.

So, your original equation can be written as

$\displaystyle \dfrac{d}{dx} \left(y' + xy' - y + xy \right) = 0$.

Your turn!