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Ordinary Differential Equations and Initial-Value Problems : Unique Solution

Find the unique solution satisfying initial conditions y(1)=0 and y'(1)=1 for
ty''-(t+2)y'=2y=0, t not equal to 0

Solve t^2y''+3ty'+5y=0

Solve the initial value problem.
y''-2y'+y=2e^t+3
y(0)=4 and y'(0)=3

Solution Preview

ty''-(t+2)y'+2y=0
=>t(y''-y')=2(y'-y)
Let x=y'-y
=>tx'=2x
=>x=y'-y=a t^2 ...

Solution Summary

An ODE and IVP are solved.

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