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Solve the initial value problems:
1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1
Answers: 2
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Solve the initial value problems:
1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1...
1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1...
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