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Why is the general solution to a inhomogeneous equation the particular solution added with the solution to the respective homogeneous equation?

Whenever I ask why I'm told to not worry about it, but I have a hard time doing something I do not understand.

Suppose I am asked to find a function y that satisfies the equation:

$$y' + 3y = 6x+5$$

My math teacher told me to first find a particular solution generalized as $ax+b$: $$y' + 3y = 6x+5$$ $$a + 3ax+3b = 6x+5$$ And thus I get an equation system: $3a = 6, a+3b = 5$ A particular solution is $2x + 1$

Now to the part I do not understand. Why do we get the general solution by adding the function that satisfies $y' + 3y = 0$?

graydad
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Arcthor
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  • See here http://mathoverflow.net/questions/124694/reference-for-a-nice-proof-of-undetermined-coefficients. The proofs might be difficult to understand (depending on your current level) which is probably why you have been told not to worry about it. – Matthew Cassell Feb 03 '15 at 16:02
  • Also, as I said to someone yesterday, the method is based upon the superposition principle which comes from physics. Because you are dealing with a linear system, the sum of the homogeneous and particular solutions gives the total solution to the differential equation. – Matthew Cassell Feb 03 '15 at 16:03

1 Answers1

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If $h(x)$ is a solution of the homogeneous equation $y'+ay=0$ and $s(x)$ a solution of $y'+ay=g(x)$ deriving the sum $h(x)+s(x)$ we get $$h'(x)+s'(x)=-ah(x)-as(x)+g(x)$$ and this shows that $h+s$ is another solution of the equation with non zero RHS. More intuitively the solutions of the homogeneous form a one dimensional vector space and the solution of the non homogeneous is an affine space with the previous vector space as a direction

marwalix
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