# Verify that the given function(explicit or implicit)is a solution of the corresponding differential equation $xy=\log\: y+C\qquad:y'=\large\frac{y^2}{1-xy}$$\;(xy\neq1) ## 1 Answer Toolbox: • When the given function is in the form product, we apply the product rule, which states differentiation of uv = uv' + vu'. Differentiation of \log y is \large\frac{1}{y}$$ . y'$
Step 1:
Differentiating on both sides we get,
On the LHS, applying the product rule we get, $xy' + y.1$ and on differentiating RHS we get $\large\frac{1}{y}$$. y' Hence xy' + y = \large\frac{1}{y }$$. y'$
Step 2:
Multiplying on both sides by y we get,
$y^2 + xyy' = y'$
taking y' as the common factor we get
$y'(xy - 1) = - y^2$
Hence $y' =\large\frac{ y^2}{1 - xy}$
Hence the answer is a solution to the given function.