Question

Determine which of the following initial value problems is correctly associated to the longest interval guaranteed by the existence and uniqueness theorem. y O [0, 4); ty" = 0, y(1) = 0, y (1) = -2 O (5,00); (x – 5)3 dy – 3(x + 2)2 dy CU 3+3 v(2) = -1,v' (2) = 1 1 d.c3 dx2 0(-1,1); 2(t– 1)y" + 3ty - y=et, y(0) = 1, y (0) = 0 (-0,3); xạy" + 2xy – y = 713, y(0) = 0. y' (0) = 1

Answer 1

Existance and uniqueness theorem : Consider on y" + p(a) zy' +9 (0) Y = of(a). y(x) = Y, ;'(X) = y. If the functien þc«) and g(x) and fou are continuous the interval I: <a<ß containing the point a = to. a unique solution 3 = 4(x) of the problem and this solution exists throughout the interval I. Then there exists

(1) Now consider t2 y" y y (1) = 0 , Y'(1)=-2 t-4 y" y SO tz (t-1) for existance and uniqueness of solution sould be continuous in the interval t() 1 Xo = 1 which contains point t to and t = 4 for to 1 be continuous te. t2(t-4) Thus also le (0,4). (0,4) is the longest interval but in our option [0,4] is given which is not correct. (ii) (x-5² g 3(x+2)?dhe 1(2) -- 29 (2)=1 (x+3) 01322 - 3(272)2 (X-5)3 px? (2-5)3(x+3) 2 20 das dx² d²v 2 29 da3 also again for solution be exist and unique x + 5 and at-3. 2 E (-3,5) Thus (-3,5) is correct interval for solution to be exist and unique. but in option (5,00) is given which is not correct.

(iii) 2(t+1)y" + 3t y'-y = et y (0) = 1, Y(0) = 4 2 y + 3t 2 (12-1) #0 y! - ? (12-1) (121) For existance and ceniqueness of solution (tt) (t+1) #0 t = 1 and t = -1 also 0ė (-1, 1). Thus Solection exists and will be unique in the interval (-1, 1). our option (-1,1) is correctly associated In Thus option (III) cis correct (iv) 22y" + 2xy' y y (0)=0, y'(o)=1 x-3 y" + 23 y x2(x-3) For Solution C existence and unique ness) x² 27o and (x-3) 07 290, 71 #3 0$ (3,00) 0¢ (0,3) 0 € (-60,0) and Thus solution does not exist Thus option is incorrect.