Verify: $x^2=2y^2\log y$, and $x^2+y^2=xy\dfrac{dx}{dy}$

Verify: x²=2y²log y, and x²+y²=xy(dx)/(dy)

Miscellaneous Exercise 6(II) · Q120

x^2=2y^2\log y

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$x^2=2y^2\log y$ (implicit). Differentiate w.r.t. $y$ (since the target relation is in $\dfrac{dx}{dy}$ ): $2x\dfrac{dx}{dy}=2\left[2y\log y+y^2\cdot\dfrac1y\right]=4y\log y+2y$ , so $\dfrac{dx}{dy}=\dfrac{2y\log y+y}{x}=\dfrac{y(2\log y+1)}{x}$ . Check the claim $x^2+y^2=xy\dfrac{dx}{dy}$ : RHS $=xy\cdot\dfrac{y(2\log y+1)}{x}=y^2(2\log y+1)=2y^ …

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Q.Verify: $x^2=2y^2\log y$ , and $x^2+y^2=xy\dfrac{dx}{dy}$

Concept understanding — General and Particular Solutions of a Differential Equation

Q.Verify: x2=2y2log⁡yx^2=2y^2\log yx2=2y2logy, and x2+y2=xydxdyx^2+y^2=xy\dfrac{dx}{dy}x2+y2=xydydx​

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Q.Verify: $x^2=2y^2\log y$ , and $x^2+y^2=xy\dfrac{dx}{dy}$

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