Simple linear differential equation

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SUMMARY

The discussion focuses on solving a simple linear differential equation of the form By'' + y' = A, where A and B are constants. The complementary solution is derived as y_C = C_1 + C_2 e^{-t/B}, with characteristic roots m = 0 and m = -1/B. The particular solution is identified as y_p = A[t + B(e^{-t/B} - 1)], with the clarification that constant solutions cannot be used for the non-homogeneous equation. The final solution combines both the complementary and particular solutions.

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  • Understanding of linear differential equations
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  • Knowledge of characteristic equations
  • Basic calculus concepts, including derivatives
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  • Study the method of undetermined coefficients for particular solutions
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Hey, I know this is easy I just can't remember how to do it.

y(0)=0

and By''+y'=A A,B constants.

So complementary solution Bm^{2}+m=0 \\ \text{ therefore } m=0, \frac{-1}{B}
therfore y_{C} = C_{1}+C_{2}e^{-t/B}
Not sure what to do for particular solution though because substituting in constant doesn't give you anything

The final answer is:

y= A[t+B(e^{-t/B}-1]

Thanks
 
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Since y = constant is a solution for the homogeneous equation, it can't be a solution yp of the NH equation. For your particular solution try yp = Ct.
 
LCKurtz said:
Since y = constant is a solution for the homogeneous equation, it can't be a solution yp of the NH equation. For your particular solution try yp = Ct.

Ah yes now I remember! Thank you very much
 

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