Mathematics

Mathematics

Problem

If yy is a function of xx, then d2ydx2+y⋅dydx=0\frac{d^2 y}{dx^2} + y \cdot \frac{dy}{dx} = 0 . If xx is a function of yy , then the equation becomes a. d2xdy2−x⋅dxdy=0\frac{d^2 x}{dy^2} - x \cdot \frac{dx}{dy} = 0 b. d2xdy2+y(dxdy)3=0\frac{d^2 x}{dy^2} + y \left( \frac{dx}{dy} \right)^3 = 0 c. d2xdy2−y(dxdy)2=0\frac{d^2 x}{dy^2} - y \left( \frac{dx}{dy} \right)^2 = 0 d. d2xdy2−x(dxdy)2=0\frac{d^2 x}{dy^2} - x \left( \frac{dx}{dy} \right)^2 = 0

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