Question

8. DETAILS ZILLDIFFEQMODAP11 8.2.013. Solve the given initial-value problem. 1 2 0 5 X' X(0) = 1 1 2 X(t) =

Answer 1

Ans. initial - Value problem १ ** [1 :7*. *•[:] "[:] Now let Az 1 you Now we have to find Eigen vector and Eigen values of this makia so that take ya [ z 0.5 IA-AI) - 0 5) (0.5-x) O 30 I (-0.5-1) (0.5-0) x(-0.5-) - 0x1 = 0 (-0.25 +2) -0 -0 (A _ 0.25) = 0 (A-0.5) (n +0.5) 20 Now Here we can see that the Eigen values (A) are 1 = -0.5, 0.5

Now we have to calculate Eigenvectors for different Eigen value (N) A-0.5 Eigenvector for A-AI) 1 *[** .:)--[: :) [* :] [::. [i :) Now Reduce this matrix by applying different Rows operations sowe get R, EJ R2 [inten inten changing 1 [: :

NOW (A-0.58) -0[1] • [:-:)N-[:] 1 X, – Xzco * =*₂ so the Eigenvectors corresponding to the eigenvalues A=0.5 is KI X2 [* X2 Now let x₂ = 1 so we get K, = [:] Now for = -0.5 Eigenvalues (A) (A-AI) find Eigenvoctors o O.S O + 0.5 1 -0.5

0.5 0 [. -0.5 0.5 :] [: [:: - 1 Now again Reduce this matrix by applying different Rows operations so we get R₂ R2- - RA 1 0 [ O 0 Now (A+0.51) X, 1 0 ][*]-[:] Xa 6 O Xi so the Eigen vectors corresponding to the Eigenvalue (0) is = -0.5

K₂ х, det x2 к, [:] Now we know that the geneual solution is X(t) a kielt + C₂ kq e lat + (+) At с, к, е + Cqkze x (t) = a 2 *[]*** - [:]-**- ***0-12:00 12:30 when t=o مم حم

“[:] *-[:] [: G 1 IX G + oxcy = 5 IxG + Ixcą = 9 2 from C +0 = 5 C = 5 Put the value of c, in equation ® so we get G + C₂ =9 s + C₂ =9 C₂ =9-5 C₂=4 Now put the values of e, and Ca in equation ® so we get x(t) = 4 + C 0.st - 6.st e e "11] -[:] * [+] ***+ "[:]*** X 1 or this can be written as

X -- [:]•* - •[:] e Ans.