Question

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Answer 1

It is given that, x 지 d dt x = 1 x= 3 4 x2 Hence, d dt. X2 31 102 da, x + 2 at la 3x, + 4x2 dt Thus, x + x2 2 dt daz - 3x + 42. las dt dai 4 + 3 Differentiating is with respect to t, We obtain, dxz dt d'az_ 3 (1 + 2) + 4 d r [ from &] dra dt dt² 7 at 4 dan at 3x2 = 3x, dt² d x2 daa_ 4x [from 2] da – 3x2 dt - 4 at dt² dne dt² 5 da + x2 at

Let x2C+) = cemt(4+0) be a trial solution, of lii) Hence, auxiliary equation is m - (-5) ± √(-5) _ 4. 1.1 m²_5 mtl=0 Hence, m 2 5 + 25 - 4 m a 2 =) m 5 + √21 2 2 Let &= 5+/21 and & 5-21 12 21