A function $$ y(t) $$ satisfies the differential equation $$ \frac{d y}{d t}=-y^{4}-3 y^{3}+54 y^{2} $$ (a) What are the constant solutions of this equation? Separate your answers by commas. (b) For what values of $$ y $$ is $$ y $$ strictly increasing? and

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**(a)** We set the right‐hand side of the differential equation equal to zero: $$ -y^4-3y^3+54y^2=0. $$ Factor out $$y^2$$: $$ y^2\bigl(-y^2-3y+54\bigr)=0. $$ This yields the solutions: $$ y^2=0 \quad \text{or} \quad -y^2-3y+54=0. $$ The first equation gives $$ y=0. $$ For the second equation, multiply through by $$-1$$ to obtain: $$ y^2+3y-54=0. $$ Factor the quadratic: $$ y^2+3y-54=(y-6)(y+9)=0, $$ which gives $$ y=6 \quad \text{or} \quad y=-9. $$ Thus, the constant solutions are: $$0, 6, -9$$. **(b)** We need to find the values of $$ y $$ for which the derivative $$ \frac{dy}{dt}=-y^4-3y^3+54y^2 $$ is positive. We already factored the expression in part (a): $$ \frac{dy}{dt}=-y^2(y^2+3y-54). $$ Recognize that $$ y^2+3y-54=(y-6)(y+9), $$ so that $$ \frac{dy}{dt}=-y^2\,(y-6)(y+9). $$ To determine when $$\frac{dy}{dt}>0$$, we analyze the sign of the product $$-y^2\,(y-6)(y+9)$$ over the different intervals determined by the zeros $$ y=-9 $$, $$ y=0 $$, and $$ y=6 $$. 1. **For $$ y < -9 $$:** - $$ y^2 > 0 $$ (always positive except at $$ y=0 $$). - $$ y-6 < 0 $$ (since any $$ y < -9 $$ is less than 6). - $$ y+9 < 0 $$ (since $$ y < -9 $$). Thus, $$(y-6)(y+9)$$ is the product of two negatives (positive), and multiplying by $$-y^2$$ gives a negative result. So, $$\frac{dy}{dt} < 0$$. 2. **For $$ -9 < y < 0 $$:** - $$ y^2 > 0 $$. - $$ y-6 < 0 $$ (since $$ y < 0 $$). - $$ y+9 > 0 $$ (since $$ y > -9 $$). Thus, $$(y-6)(y+9)$$ is negative, and the negative multiplier makes the whole expression positive: $$\frac{dy}{dt}>0$$. 3. **For $$ 0 < y < 6 $$:** - $$ y^2 > 0 $$. - $$ y-6 < 0 $$ (since $$ y < 6 $$). - $$ y+9 > 0 $$. Again, $$(y-6)(y+9)$$ is negative and the overall expression becomes positive: $$\frac{dy}{dt}>0$$. 4. **For $$ y > 6 $$:** - $$ y^2 > 0 $$. - $$ y-6 > 0 $$. - $$ y+9 > 0 $$. Now, $$(y-6)(y+9)$$ is positive, and the multiplier $$-y^2$$ makes the whole expression negative: $$\frac{dy}{dt}<0$$. Thus, the function satisfies $$\frac{dy}{dt}>0$$ (i.e., $$y$$ is strictly increasing) for $$ -9 < y < 0 \quad \text{and} \quad 0 < y < 6. $$ **Answer:** (a) $$0, 6, -9$$ (b) $$-9**(a)** The constant solutions are $$0, 6, -9$$. **(b)** The function $$ y(t) $$ is strictly increasing when $$ -9 < y < 0 $$ and $$ 0 < y < 6 $$.

A function satisfies the differential equation

(a) What are the constant solutions of this equation?
Separate your answers by commas.
(b) For what values of is strictly increasing?
and

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A function satisfies the differential equation (a) What are the constant solutions of this equation? Separate your answers by commas. (b) For what values of is strictly increasing? and