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I am in need of explinations on how to resolve second degree linear equations with a constant.

I had a maths test last saturday and this was the last question :

Resolve y"+3y'+2=cos(t)

Having almost fallen into the trap of resolving y"+3y'+2y=cos(t), I passed the 2 over by cos(t) and tried to resolve y"+3y'=cos(t)-2, however I doubt that that's possible (maybe I'm wrong).

Anyhow I am looking for some help in resolving this equation because I have the same equation given as a test tomorrow morning.

Many thanks,

Callum

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# Explinations needed, second degree linear equations with a constant

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