MATH 238 Lecture Notes - Lecture 5: Fundamental Solution, Wronskian
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Diff eq homework 5: (cid:884)y -5y"+(cid:884)y=(cid:882) ; y(cid:523)(cid:882)(cid:524)= -(cid:883), y"(cid:523)(cid:882)(cid:524) = -5, find the general solution of the differential equation. Replace y" by m and y by m2 where m= , m2 = 22. Therefore the general solution y= a (e2t) +be0. 5t ------(1) M = 2, 0. 5: impose the initial conditions to obtain the unique solution of the initial value problem. Y1(t)= cos(t: verify that the given functions are solutions to the differential. Y(cid:883) (cid:523)t(cid:524) = -cos(t: y +y= (cid:882) ; y1(t) = cos(t), y2(t) = sin(t) This is not true for all values of t; therefore y1(t) = cos(t) is not a solution of (1) substitute the above values in (1) 0=0, this is true for all values of t. Now let y2(t)= sin(t) therefore y2(t) = sin(t) is a solution of (1); hence it is a solution of the. Y1(t), y2(t) form two fundamental solution wronskian (y1,y2) = Given that y1(t) = cos(t), y2 = sin (t)
