Example:(X^2D^2-xD-15)y=0,y(1)=0.1,y'(1)=-4.5

tricotasu 2020-12-09 Answered
Example:(X2D2xD15)y=0,y(1)=0.1,y(1)=4.5
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Expert Answer

Theodore Schwartz
Answered 2020-12-10 Author has 99 answers
Here the given differential equation is
x2yxy15y=0,y(0)=0.1,y(0)=4.5
Let lnx=z then
dydx=dydzdzdx=1xdydz.
similary
x2d2ydx2=d2ydz2dydz.
Using above two expressions, the original differential equation reduces to
(d2ydz2dydz)dydz15y=0
d2ydz22dydz15y=0.
The auxiliary equation of the differential equation d2ydz22dydz15y=0 becomes
m22m15=0
(m5)(m+3)=0
m=5,3.
Therefore the general solution of the differential equation is
y(x)=ae5z+b3z=ax5+bx3.
Here we can not put y(0)=0.1 since in the general solution term bx3 is present.
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