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Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem.

xx+1y'''-3xy'+y=0y-12=1,  y'-12=y''-12=0

Short Answer

Expert verified

Hence, the largest interval for the existence of a unique solution on (a, b) to the given initial value problem is-1,  0.

Step by step solution

01

Step 1:Solve the given equation

The given equation isxx+1y'''-3xy'+y=0.

Divide both sides by x(x+1) in the above equation,

y'''-3x1xx+1y'+1xx+1y=0

Simplify the above equation,

y'''-3x+1y'+1xx+1y=0

Compare with the standard form of a linear differential equation,

y'''+pxy''+qxy'+rxy=sx

One has,qx=-3x+1,  rx=1xx+1

02

Step 2:Check the continuity

qx=-3x+1 is continuous for all x≠-1.

rx=1xx+1is continuous in x≠0, -1.

03

Step 3:The largest interval (a, b)

Now q and r continuous for allx∈-∞,  -1∪-1,  0∪0,  ∞

And the initial condition is defined atx0=-12

And-12∈-1,  0

Hence, the largest interval for the existence of a unique solution on (a, b) to the given initial value problem is-1,  0.

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