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Exercise 6.4 · Q64

Q.Solve: x dy+2y⋅dx=0x\,dy+2y\cdot dx=0, when x=2, y=1x=2,\ y=1.

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x dy+2y dx=0x\,dy+2y\,dx=0 is directly separable: dyy=−2dxx\dfrac{dy}{y}=-2\dfrac{dx}{x}. Integrating: log⁡y=−2log⁡x+c1\log y=-2\log x+c_1, i.e. x2y=cx^2y=c. …

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