Question

Solve the differential equation \[({x^2} + {y^2})dx - 2xydy = 0\].

Answer 1

Hint: Determine \[\dfrac{{dy}}{{dx}}\] from \[({x^2} + {y^2})dx - 2xydy = 0\] and observe it is a function of \[\dfrac{y}{x}\]. Substitute another variable in terms of \[\dfrac{y}{x}\] and solve the differential equation and finally convert it in terms of y and x.

Complete step-by-step answer:
A differential equation is an equation that relates one or more functions and their derivatives.
We are given the expression \[({x^2} + {y^2})dx - 2xydy = 0\] and we need to solve for this differential equation.
As a first step, we need to find \[\dfrac{{dy}}{{dx}}\] from \[({x^2} + {y^2})dx - 2xydy = 0\].
\[({x^2} + {y^2})dx - 2xydy = 0\]
Taking -2xydy to the right-hand side of the equation, we get:
\[({x^2} + {y^2})dx = 2xydy\]
Dividing both sides of the equation by dx, we obtain as follows:
\[{x^2} + {y^2} = 2xy\dfrac{{dy}}{{dx}}\]
Solving for \[\dfrac{{dy}}{{dx}}\] in terms of x and y, we get:
\[\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} + {y^2}}}{{2xy}}\]

\[\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} + {y^2}}}{{2xy}}................(1)\]
Let \[F(x,y) = \dfrac{{{x^2} + {y^2}}}{{2xy}}\], we now find \[F(\lambda x,\lambda y)\].
\[F(\lambda x,\lambda y) = \dfrac{{{{(\lambda x)}^2} + {{(\lambda y)}^2}}}{{2(\lambda x)(\lambda y)}}\]
\[F(\lambda x,\lambda y) = \dfrac{{{\lambda ^2}{x^2} + {\lambda ^2}{y^2}}}{{2{\lambda ^2}xy}}\]
\[F(\lambda x,\lambda y) = \dfrac{{{x^2} + {y^2}}}{{2xy}}\]
\[F(\lambda x,\lambda y) = F(x,y)\]
Hence, F(x, y) is a homogeneous differential equation.
Therefore, we make the substitution y=ux.
\[u = \dfrac{y}{x}.............(2)\]
\[y = ux\]
Differentiating both sides with respect to x, we get as follows:
\[\dfrac{{dy}}{{dx}} = x\dfrac{{du}}{{dx}} + u.............(3)\]
From equation (1), we have:
\[\dfrac{{dy}}{{dx}} = \dfrac{x}{{2y}} + \dfrac{y}{{2x}}\]
Substituting equation (2) and equation (3) into the above equation, we have:
\[x\dfrac{{du}}{{dx}} + u = \dfrac{1}{{2u}} + \dfrac{u}{2}\]
Taking u to the right-hand side of the equation and simplifying, we have:
\[x\dfrac{{du}}{{dx}} = \dfrac{1}{{2u}} + \dfrac{u}{2} - u\]
\[x\dfrac{{du}}{{dx}} = \dfrac{1}{{2u}} - \dfrac{u}{2}\]
Simplifying the fractions, we have:
\[x\dfrac{{du}}{{dx}} = \dfrac{{1 - {u^2}}}{{2u}}\]
Now, we can solve this differential equation using separation of variables method.
Taking all terms with u to the left-hand and taking all terms with x to the right-hand side, we have:
\[\dfrac{{2u}}{{1 - {u^2}}}du = \dfrac{{dx}}{x}\]
Integrating both sides, we have:
\[\int {\dfrac{{2u}}{{1 - {u^2}}}du} = \int {\dfrac{{dx}}{x}} \]
Let \[1 - {u^2} = t\].
\[1 - {u^2} = t\]
Differentiating both sides we have:
\[ - 2udu = dt\]
Then the integral is given as follows:
\[ - \int {\dfrac{1}{t}dt} = \int {\dfrac{{dx}}{x}} \]
Integration of \[\dfrac{1}{x}\] is \[\log \left| x \right|\].
\[ - \log \left| t \right| = \log \left| x \right| + C\]
Simplifying, we get:
\[ - \log \left| t \right| - \log \left| x \right| = C\]
\[\log \left| {xt} \right| = - C\]
We know that \[1 - {u^2} = t\].
\[\log \left| {x(1 - {u^2})} \right| = - C\]
We know that \[u = \dfrac{y}{x}\], then, we have:
\[\log \left| {x\left( {1 - {{\left( {\dfrac{y}{x}} \right)}^2}} \right)} \right| = - C\]
Simplifying, we get:
\[\log \left| {x\left( {{{\dfrac{{{x^2} - y}}{{{x^2}}}}^2}} \right)} \right| = - C\]
\[\log \left| {{{\dfrac{{{x^2} - y}}{x}}^2}} \right| = - C\]
Let \[ - C = \log c\], then we have:
\[\log \left| {{{\dfrac{{{x^2} - y}}{x}}^2}} \right| = \log c\]
\[\left| {{{\dfrac{{{x^2} - y}}{x}}^2}} \right| = c\]
\[\left| {{x^2} - {y^2}} \right| = c|x|\]
Hence, the solution to the given differential equation is \[\left| {{x^2} - {y^2}} \right| = c|x|\].

Note: If you use the integration of \[\dfrac{1}{x}\] as just \[\log x\], you will end up in a wrong answer. Integration of \[\dfrac{1}{x}\] is \[\log \left| x \right|\].