The constant solution of $\frac { d x } { d t } = 2 x ^ { 2 } + y ^ { 2 } - 1 , \frac { d y } { d t } = x - 2 y$ are
A) $x = 2 / 3 , y = 1 / 3$
B) $x = 2 / 3 , y = 1 / 3$ and $x = - 2 / 3 , y = - 1 / 3$
C) $x = - 2 / 3 , y = - 1 / 3$
D) $x = 2 / \sqrt { 7 } , y = 1 / \sqrt { 7 }$
E) $x = 2 / \sqrt { 7 } , y = 1 / \sqrt { 7 }$ and $x = - 2 / \sqrt { 7 } , y = - 1 / \sqrt { 7 }$

Question

Quizplus · Accepted Answer

For constant solutions, both derivatives must be zero. Setting $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 0$, we solve the system of equations $2x^2 + y^2 - 1 = 0$ and $x - 2y = 0$. Solving these simultaneously gives two sets of solutions: $x = 2/3, y = 1/3$ and $x = -2/3, y = -1/3$.

The Constant Solution Of dxdt=2x2+y2−1,dydt=x−2y\frac { d x } { d t } = 2 x ^ { 2 } + y ^ { 2 } - 1 , \frac { d y } { d t } = x - 2 ydtdx​=2x2+y2−1,dtdy​=x−2y

The Constant Solution Of $\frac { d x } { d t } = 2 x ^ { 2 } + y ^ { 2 } - 1 , \frac { d y } { d t } = x - 2 y$