Question

How do you find $ \dfrac{{dy}}{{dx}} $ for the equation $ {y^2} - xy = - 5 $ ?

Answer 1

Hint: In order to determine the first order derivative of the equation $ {y^2} - xy = - 5 $ , we will consider the equation and differentiate the equation with respect to $ x $ and use the differential formulas like $ \dfrac{{dy}}{{dx}}\left( {xy} \right) = x\dfrac{{dy}}{{dx}} + 1 $ , $ \dfrac{{dy}}{{dx}}\left( C \right) = 0 $ . And evaluate to determine $ \dfrac{{dy}}{{dx}} $.

Complete step-by-step answer:
Now, we need to determine $ \dfrac{{dy}}{{dx}} $ for the equation $ {y^2} - xy = - 5 $ .
The given equation is $ {y^2} - xy = - 5 $ $ \to \left( 1 \right) $
Now, let us differentiate equation $ \left( 1 \right) $ with respect to $ x $ ,
 $ {y^2} - xy = - 5 $
Differentiation of $ xy $ is given by,
 $ \dfrac{{dy}}{{dx}}\left( {xy} \right) = x\dfrac{{dx}}{{dx}} + \dfrac{{dx}}{{dy}}y $
 $ = x\dfrac{{dy}}{{dx}} + \left( 1 \right)y $
 $ = x\dfrac{{dy}}{{dx}} + y $
And, the differentiation of a constant term is always zero, i.e., $ \dfrac{{dy}}{{dx}}\left( C \right) = 0 $ .
Hence, applying these formulas, we have,
 $ 2y\dfrac{{dy}}{{dx}} - y - x\dfrac{{dy}}{{dx}} = 0 $
Let us bring $ \dfrac{{dy}}{{dx}} $ to one side of the equation, and then we have,
 $ 2y\dfrac{{dy}}{{dx}} - x\dfrac{{dy}}{{dx}} = y $
 $ \dfrac{{dy}}{{dx}}\left( {2y - x} \right) = y $
 $ \dfrac{{dy}}{{dx}} = \dfrac{y}{{2y - x}} $
Therefore, $ \dfrac{{dy}}{{dx}} $ for the equation $ {y^2} - xy = - 5 $ is $ \dfrac{y}{{2y - x}} $.
So, the correct answer is “$ \dfrac{y}{{2y - x}} $”.

Note: A differential equation is an equation with a function and one or more of its derivatives or differentials.he order of the differential equation is the order of the highest order derivative present in the equation. The degree of the differential equation is the power of the highest order derivative, where the original equation is represented in the form of a polynomial equation in derivatives such as $ \dfrac{{dy}}{{dx}},\,\dfrac{{{d^2}y}}{{d{x^2}}},\,\dfrac{{{d^3}y}}{{d{x^3}}} \ldots $