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 $\log \left| x \right|$.