Solve the differential equation. $$y ^ { t } = \frac { - 2 x } { y }$$
A) $y ^ { 2 } = 2 x ^ { 2 } + C$
B) $2 \ln y = - 2 x ^ { 2 } + C$
C) $2 \ln y = 2 x ^ { 3 } + C$
D) $y ^ { 2 } = - 2 x ^ { 2 } + C$
E) $y ^ { 2 } = - 2 x ^ { 3 } + C$

Question

Quizplus · Accepted Answer

We can rewrite the given equation as $y\,dy=-2x\,dx$. Integrating both sides, we get $\frac{y^2}{2}=-x^2+C$, where $C$ is a constant of integration. Solving for $y$, we get $y=\pm\sqrt{2(-x^2+C)}=\pm\sqrt{2} \sqrt{-x^2+C}$. Since $y$ can be either positive or negative, we include both possibilities in our solution. Hence, the answer is $y ^ { 2 } = - 2 x ^ { 2 } + C$.

Solve the Differential Equation yt=−2xyy ^ { t } = \frac { - 2 x } { y }yt=y−2x​

Solve the Differential Equation $y ^ { t } = \frac { - 2 x } { y }$