Check that the provided functions serve as a foundation for the solutions to the provided...

beljuA

beljuA

Answered

2021-01-04

Check that the provided functions serve as a foundation for the solutions to the provided equation and resolve the provided initial value problem.
4x23y=0, y(1)=3, y(1)=2.5, the basis of solution are y1=x12 and y2=x(32)

Answer & Explanation

pattererX

pattererX

Expert

2021-01-05Added 95 answers

Given:
4x23y=0,y(1)=3,y(1)=2.5, the basis of solution are y1=x12 and y2=x(32)
4x2y3y=0
Let x=et
dxdt=et
dydx=dydtdxdt=etdydt
etdydx=dydt
xdydx=dydt
Similarly x2d2ydx2=d2ydt2dydt
4x2y3y=0
4d2ydt24dydt3y=0
Auxiliary equation is given by:
4m24m3=0
m=32,12
y(t)=c1e32t+c2e12t
put t=lnx  (x=et)
y(x)=c1x32+c2x12
Consequently, every solution is a linear combination of
x32 and x12
also x32 and x12 are linearly independent so [x32,x12] forms a basis for the solution of 4x2y3y=0
Now, y(x)=c1x32+c2x12
given y(1)=3, y(1)=25
y(1)=3
3=c1+c2
y(x)=32c1x1212c2x32
y(x)=2.5
2.5=32c112c2
5=3c1c2
Solving we get:
c1=2, c2=1
y(x)=2x32+x12

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