Differential equations MOC

Solving non-homogenous second order ODEs

For any second-order ODE

𝐿[𝑦]=𝑦″+𝑝(𝑡)𝑦′+𝑞(𝑡)𝑦=𝑔(𝑡)

it is clear that the difference of any two solutions solves the related homogenous ODE. Namely, given 𝐿[𝑌1] =𝑔(𝑡) and 𝐿[𝑌2] =𝑔(𝑡), it is clear that

𝐿[𝑌1−𝑌2]=𝐿[𝑌1]−𝐿[𝑌2]=𝑔(𝑡)−𝑔(𝑡)=0

It follows that given the general solution to the related homogenous ODE 𝑦𝑐(𝑡), called the complimentary solution, and any one particular solution to the full ODE 𝑦𝑝, then

𝑦=𝑦𝑐+𝑦𝑝=𝑐1𝑦1+𝑐2𝑦2+𝑦𝑝

since

𝐿[𝑦𝑐+𝑦𝑝]=𝐿[𝑦𝑐]+𝐿[𝑦𝑝]=𝑔(𝑡)

Finding a particular solution

In practice, a variety of methods may be used to find a particular solution once 𝑦𝑐 has been found

• Method of undetermined coëfficients (uses an Ansatz)
• Method of annihilation
• Method of variation of parameters (similar to Reduction of order (homogenous second-order differential equation))

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