Mathematics, 18.07.2019 11:30 anonymous777739
Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8
Answers: 1
, so that
![\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)](/tpl/images/0103/8646/3c629.png)
with roots at 
. This means the characteristic solution for 
is
, this is
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Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8...
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