Question

If $\log \sqrt {{x^2} + {y^2}} = {\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)$, then prove that $\dfrac{{dy}}{{dx}} = \dfrac{{x + y}}{{x - y}}$.

Answer 1

Hint: Use the basic rule of logarithm to simplify the given equation and then differentiate the given equation with respect to $x$ on both sides to obtain the differential equation. Also, Use the formula of differentiation of tan inverse $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$.

Complete step-by-step solution:
We know that the solution of the differential equation is given in the questions is $\log \sqrt {{x^2} + {y^2}} = {\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)$.
We have to obtain the differential equation from the given solution. First, we rewrite the given equation in a simpler form, we have
\[
\Rightarrow \log {\left( {{x^2} + {y^2}} \right)^{1/2}} = {\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)\\
\Rightarrow \dfrac{1}{2}\log \left( {{x^2} + {y^2}} \right) = {\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right)
\]
Now, we differentiate the above equation with respect to $x$, we get,
\[
\Rightarrow \dfrac{d}{{dx}}\left\{ {\log \left( {{x^2} + {y^2}} \right)} \right\} = \dfrac{{dy}}{{dx}}\left\{ {{{\tan }^{ - 1}}\left( {\dfrac{x}{y}} \right)} \right\}\\
\Rightarrow \dfrac{1}{2}\left( {\dfrac{1}{{{x^2} + {y^2}}}} \right)\dfrac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \dfrac{1}{{1 + {{\left( {y/x} \right)}^2}}}\dfrac{d}{{dx}}\left( {\dfrac{y}{x}} \right)\\
\Rightarrow \dfrac{1}{{2\left( {{x^2} + {y^2}} \right)}}\left( {2x + 2y\dfrac{{dy}}{{dx}}} \right) = \dfrac{{{x^2}}}{{\left( {{x^2} + {y^2}} \right)}}\left\{ {\dfrac{{x \cdot \dfrac{{dy}}{{dx}} - y\dfrac{{dx}}{{dx}}}}{{{x^2}}}} \right\}
\]
In the above equation \[\dfrac{1}{{\left( {{x^2} + {y^2}} \right)}}\] is common on both sides of the equation, which can be cancelled out from both sides of the equation.
After simplifying the above equation into a simpler form, we have,
\[
\Rightarrow x + y\dfrac{{dy}}{{dx}} = x \cdot \dfrac{{dy}}{{dx}} - y\dfrac{{dx}}{{dx}}\\
\Rightarrow \dfrac{{dy}}{{dx}}\left( {y - x} \right) = - \left( {y + x} \right)\\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \left( {y + x} \right)}}{{\left( {y - x} \right)}}
\]

Now we rewrite the above equation by dividing minus on numerator and denominator, we have,
$\Rightarrow$ \[\dfrac{{dy}}{{dx}} = \dfrac{{x + y}}{{x - y}}\]

Hence, the given equation is the solution of the differential equation \[\dfrac{{dy}}{{dx}} = \dfrac{{x + y}}{{x - y}}\].

Note: The differential equation is a type of equation that contains differential coefficients, derivatives or differentials of one or more than one dependent variable with respect to one or more than one independent variables. The different types of differential equations are ordinary differential equations and partial differential equations. The solution of a given differential equation is the relationship between the dependent and independent variable that completely satisfies the differential equation.