Find the Solution (1+1?) ds + 21[st²– 3(1+1²)²jdt = 0; when t = 0, s = 2. A t = (1 + s²)[3-e³} B s = (1 - 12)[2-e- c t = (1 – s²)[2– e ~s*1 s = (1 + 12)[3- e-ů

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