Putting, x=0 in y2+loge(cos2x)=y we get y=0,1
2y⋅y′+cos2x1⋅2cosx(−sinx)=y′
⇒2y⋅y′−2tanx=y′……(1)
y′(0)=0 for y=0&y=1.
Differentiating (1), 2y⋅y′′+2(y′)2−2sec2x=y′′,
y′′(0)=−2 for y=0
y′′(0)=2, for y=1
∴∣y′′(0)∣=2
If y2+loge(cos2x)=y,x∈(−2π,2π) then :
Held on 3 Sept 2020 · Verified 6 Jul 2026.
y′′(0)=0
∣y′(0)∣+∣y"(0)∣=1
∣y"(0)∣=2
∣y′(0)∣+∣y"(0)∣=3
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