# Solution Of Mathematical Methods By S.m Yousuf S

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. = e−t has characteristic equation k − 1 = 0 with solution k =The solution to the homogeneous equation is therefore (with arbitrary/free.. Therefore, let xp = Be−t be a solution to the non-homogeneous t equation (particular solution). Substitution gives: Be−(t+1) − Be. ⇔ e 1 = . B = −1 e −1 1−e The general solution is accordingly: xt = xh + xp = A + ( t t e )e.instructor solution manual for mathematical methods the physical

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there exists an orthogonal matrix P such that Pt AP = D, where D is a diagonal matrix made up from the eigenvalues of A and we are told that all of the eigenvalues and eigenvectors of this matrix are real. The orthogonal matrix P has column vectors given by the orthonormal set of vectors S = {x1 , x2 , , xn } where xi is the eigenvector corresponding to the eigenvalue λi , i.e. Axi = λi xi . As such, we have PPt = I and so, writing this out in full we have: —— xt —— 1 —— xt —— 2 I = .**Suggested**

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