Merry Christmas and a Happy New Year
Does the following differential equation not have a solution for A, and is B = -1 in the general solution ?
Also does the general solution of the following differential equation look anything like what I've given?
gen eqn: , where
Thank you all for your help.
If , then
The above equation is easily solvable using the "integrating factor" method.
I think . This means
The above equation is easily solvable, again, using the "integrating factor" method.
A general tip:
While checking, if any obtained answer to an equation is the solution to the original question, you can always help yourself by substituting the answer back and verifying whether the answer satisfies the equation.
Moreover one should always do this(subbing back) while solving equations, since extraneous roots could have crept in at some steps.
Here's a silly example that makes its point:
Solve: x = 1:
But observe that x = -1 doesnt satisfy x = 1. This happened because the second step is not reversible
Thus even in exams you can sub back and check, and be confident about your answer. Note that this tip is not just for differential equations, but any kind of equation.
Why depend on others when you can help yourself, right?
The idea of seeking a particular solution and reducing the problem to one that is homogeneous I think is easier for most.
The solution is
Next question, is the complimentary solution a part of the nonhomogeneous term. In this case yes! To use the method of undetermined coefficients it is necessary to "bump up" the solution and seek a particluar solution of the form
And you're expected to know how to construct a particular solution. Have you reviewed your class notes and textbook?
Hey, i got a few questions sorry
1 - I haven't come across the term "Bump Up" in all my notes and texts, can you explain?
2 -WELL I've tried to do it again, here's what I've got. PLEASE tell me where I've gone WRONG
The characteristic equation:
The roots to this equation: and
therefore the complementary function:
therefore my general solution is
it has powers and exponential's Therefore one would try a guess of
and substituting gives
Expanding gives (yes, the B's cancel but why?) The reason is that is part of the complimentary solution so
identically! So in this case, what do we do for a guess. Let us onsider the following first order example
The complimentary equation is
the characteristic equations is
the complimentary solution is
However, since eqn. (1) is linear we solve giving
In this solution we have two parts, the first the complimentary and the second the particular. We notice the extra x so in order to guess a particular solution when rhs contains the complimentary solution we "bump" up the exponential in our first guess and use
and when we expand we get
from which we obtain and the particular solution