91影视

An explicit solution is one in which the dependent variable is expressed directly in terms of the independent variable and constants.

Let the expression be\( - 2{x^2}y + {y^2} = 1\).

Divide \(dx\)on both sides of the equation.

\(\begin{array}{c}\frac{{(2xy)dx}}{{dx}} + \frac{{\left( {{x^2} - y} \right)dy}}{{dx}} = \frac{0}{{dx}}\\2xy + \left( {{x^2} - y} \right)\frac{{dy}}{{dx}} = 0\end{array}\)

Let the first derivative of the above function is

\(\begin{aligned}{c}\frac{d}{{dx}}\left( { - 2{x^2}y + {y^2}} \right) &= \frac{d}{{dx}}(1)\\ - 4xy + y'\left( { - 2{x^2}} \right) + 2yy' &= 0\\y' &= \frac{{4xy}}{{ - 2{x^2} + 2y}}\end{aligned}\)

Substitute \(y\) and \(y'\) into the left-hand side of the differential equation.

\(\begin{aligned}{c}2xy + \left( {{x^2} - y} \right)\frac{{4xy}}{{ - 2{x^2} + 2y}} &= 0\\2xy + \frac{{\left( {{x^2} - y} \right)4xy}}{{ - 2\left( {{x^2} - y} \right)}} &= 0\\2xy - 2xy &= 0\\0 &= 0\end{aligned}\)

That is same as the right-hand side of the differential equation. The indicated function is an explicit solution of the given differential equation.

Hence the interval of the solution is \(( - \infty ,\infty )\).

Solve the implicit function.

\(\begin{aligned}{c} - 2{x^2}y + {y^2} &= 1\\{y^2} - 2{x^2}y - 1 &= 0\end{aligned}\)

Using the quadratic formula, \(y = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\).

Here, \(a = 1;b = - 2{x^2};c = - 1\).

\(y = {x^2} + \sqrt {{x^4} + 1} \)

Let the graph of the expression be,