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31-diffeq

# 31-diffeq - DIFFERENTIAL EQUATIONS Homework section 9.3...

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DIFFERENTIAL EQUATIONS, Math 21b, O. Knill Homework: section 9.3: 6,28,34,42,44,16*,20* LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS. Df = T f = f 0 is a linear map on the space of smooth functions C . If p ( x ) = a 0 + a 1 x + ... + a n x n is a polynomial, then p ( D ) = a 0 + a 1 D + ... + a n D n is a linear map on C ( R ) too. We will see here how to ±nd the general solution of p ( D ) f = g . EXAMPLE. For p ( x ) = x 2 - x + 6 and g ( x ) = cos( x ) the problem p ( D ) f = g is the di²erential equation f 00 ( x ) - f 0 ( x ) - 6 f ( x ) = cos( x ). It has the solution c 1 e - 2 x + c 2 e 3 x - (sin( x ) + 7 cos( x )) / 50, where c 1 , c 2 are arbitrary constants. How do we ±nd these solutions? THE IDEA. In general, a di²erential equation p ( D ) f = g has many solution. For example, for p ( D ) = D 3 , the equation D 3 f = 0 has solutions ( c 0 + c 1 x + c 2 x 2 ). The constants come from integrating three times. Integrating means applying D - 1 but since D has as a kernel the constant functions, integration gives a one dimensional space of anti-derivatives (we can add a constant to the result and still have an anti-derivative). In order to solve D 3 f = g , we integrate g three times. We will generalize this idea by writing T = p ( D ) as a product of simpler transformations which we can invert. These simpler transformations have the form ( D - λ ) f = g . FINDING THE KERNEL OF A POLYNOMIAL IN D. How do we ±nd a basis for the kernel of T f = f 00 +2 f 0 + f ? The linear map T can be written as a polynomial in D which means T = D 2 - D - 2 = ( D + 1)( D - 2). The kernel of T contains the kernel of D - 2 which is one-dimensional and spanned by f 1 = e 2 x . The kernel of T = ( D - 2)( D + 1) also contains the kernel of D + 1 which is spanned by f 2 = e - x . The kernel of T is therefore two dimensional and spanned by e 2 x and e - x . THEOREM: If
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