Differential Equations Lecture Work Solutions 14

Differential Equations Lecture Work Solutions 14 - 1. a. X...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
1. a. X 0 + λX =0 X (0) = 0 X ( π )=0 Try e rx . As we know from ODEs, this leads to the characteristic equation for r r 2 + λ =0 Or r = ± λ We now consider three cases depending on the sign of λ Case 1: λ< 0 In this case r is the square root of a positive number and thus we have two real roots. In this case the solution is a linear combination of two real exponentials X = A 1 e λx + B 1 e λx It is well known that the solution can also be written as a combination of hyperbolic sine and cosine, i.e. X = A 2 cosh λx + B 2 sinh λx The other two forms are may be less known, but easily proven. The solution can be written as a shifted hyperbolic cosine (sine). The proof is straight forward by using the formula for cosh( a + b )(s inh( a + b )) X = A 3 cosh λx + B 3 Or X = A 4 sinh λx + B 4
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online