Determine if the given function y = f(x) is a solution of the accompanying differential equation.
-$\begin{array} { l }
y ^ { \prime } = - y ^ { 2 } \
y = \frac { 1 } { x + 6 }
\end{array}$

Question

Determine if the given function y = f(x) is a solution of the accompanying differential equation.
-$\begin{array} { l } 
y ^ { \prime } = - y ^ { 2 } \
y = \frac { 1 } { x + 6 }
\end{array}$

Quizplus · Accepted Answer

We can verify that the given function y = f(x) = $\frac{1}{x+6}$ satisfies the differential equation $y' = -y^2$.

Taking the derivative of y with respect to x, we get:

$$y' = \frac{d}{dx} \left(\frac{1}{x+6}\right) = -\frac{1}{(x+6)^2}$$

Substituting this into the differential equation, we get:

$$y' = -y^2 \implies -\frac{1}{(x+6)^2} = -\left(\frac{1}{x+6}\right)^2$$

Simplifying this expression, we see that it is true for all x:

$$\frac{1}{(x+6)^2} = \frac{1}{(x+6)^2}$$

Therefore, the function y = f(x) = $\frac{1}{x+6}$ is a solution of the differential equation $y' = -y^2$.

Determine If the Given Function Y = F(x) Is a Solution