This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential equation. (a) Given the equation
y″ + py′ + qy = 0,
explain why the value of y″(a) is determined by the values of y(a) and y′(a). (b) Prove that the equation
y″ − 2y′ − 5y = 0
has a solution satisfying the conditions
y(0) = 1, y′(0) = 0, and y″(0) = C
if and only if C = 5.
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