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For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation $( iv)\quad x^2=2y^2 \log y : \qquad (x^2+y^2)\large\frac{dy}{dx}$$-xy=0$

This is fourth part of multipart q2

1 Answer

  • $\large\frac{d(e^x)}{dx }$$= e^x$; $\large\frac{d(x)}{dx}$$ = 1$; $\large\frac{d(constant\; term )} {dx}= $$0$. The product rule of differentiation states $uv = uv' + vu'$
Step 1:
Given $x^2 = 2y^2\log y$
differentiating on both sides we get,
$2x = 4y\big(\frac{dy}{dx}\big)\log y + 2y^2(\large\frac{1}{y})$
dividing throughout by 2 we get,
$x = 2y(\large\frac{dy}{dx})$$\log y +y$
$\large\frac{dy}{dx }=\large\frac{ x}{y(2\log y+1)}$
Step 2:
Substituting for $\large\frac{dy}{dx}$ in the given equation we get,
$(x^2+y^2) [\large\frac{x}{y(2\log y+1)}]$$ - xy$
subtituting for $x^2$ we get
$[2y^2\log y + y^2]\large\frac{x}{y(2\log y + 1)}$$ - xy = y^2[2\log y +1]\large\frac{x}{2\log y+1} $$- xy$
Hence the given function is a solution to the corresponding differential equation.
answered Jul 30, 2013 by sreemathi.v

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