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(i) The degree of the differential Equation \(\frac{d^2y}{dx^2}\)+cos\((\frac{dy}{dx})\)=0 is(a) 2(b) 1(c) 0(d) Not defined(ii) Solve \(\frac{dy}{dx}\) + 2y tanx = sinx; y = 0, x = \(\frac{\pi}{3}\) |
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Answer» (i) (d) Not defined. (ii) \(\frac{dy}{dx}\) + 2y tanx = sinx Then, P = 2tanx, Q = sinx IF = e∫Pdx = e∫2tanxdx = e2log sec x = sec2 x Solution is; y × IF = ∫Q(IF)dx + c ⇒ ysec2 x = ∫sinx sec2 xdx + c ⇒ ysec2 x = ∫tanx secx dx + c ⇒ ysec2x = secx + c Here; y = 0, x = \(\frac{\pi}{3}\) ⇒ 0 × sec2 \(\frac{\pi}{3}\)= sec \(\frac{\pi}{3}\) + c ⇒ c = -2 ⇒ ysec2 x = secx – 2. |
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