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Mark44

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Solving Homogeneous Linear ODEs using Annihilators

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Mark44

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Solving Homogeneous Linear ODEs using Annihilators

Continue reading the Original PF Insights Post.

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ElijahRockers

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What am I missing?

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Mark44

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The differential equation in operator notation is closely related to the characteristic equation. With homogeneous diff. equations, there isn't much point, but you can use annihilators to convert

What am I missing?

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Thank you very much for your artical. Please teach about "Series Solutions‐ Frobenius‘ Method".

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Mark44

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Thank you for spotting this. I have fixed this typo as well as another in that section.

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My question is suppose we do not make any assumptions (that the solution is in the form y = e^(rt)), using the annihilator method, how do I find the general solution?

For example, in example 1, by assuming the solution takes the form of e^(rt), we can solve for r since r is a number. But without making this assumption, how do i get the solution y = c e^(3t) from (D-3)y = 0?

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You need to take into consideration the homogenous part of the equation. Let's take y=y

My question is suppose we do not make any assumptions (that the solution is in the form y = e^(rt)), using the annihilator method, how do I find the general solution?

For example, in example 1, by assuming the solution takes the form of e^(rt), we can solve for r since r is a number. But without making this assumption, how do i get the solution y = c e^(3t) from (D-3)y = 0?

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Mark44

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If you are working with a linear, constant coefficient, homogeneous differential equation, it is reasonable to assume that the solution has the form y = ceFor example, in example 1, by assuming the solution takes the form of e^(rt), we can solve for r since r is a number. But without making this assumption, how do i get the solution y = c e^(3t) from (D-3)y = 0?

(D - 3)y = 0

is the same as Dy - 3y = 0, or equivalently, y' = 3y, where y' means ##\frac{dy}{dt}##.

The DE y' = 3y is separable, and can be solved fairly easily to get the general solution: y(t) = Ce

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