Is the constant function $y\left(t\right)=-4$ a solution to the differential equation $y^{\prime }=t^2(y+4)$?

• A. No, because the derivative $y^{\prime }$ is not zero when $y\left(t\right)=-4$.
• B. No, because $y\left(t\right)=-4$ does not satisfy the initial condition $y\left(0\right)=0$.
• C. Yes, because any constant function is a solution to this differential equation.
• D. Yes, because substituting $y\left(t\right)=-4$ into the equation makes both sides equal.

Answer: (D)

Verified step by step guidance

1. Step 1: Understand the problem. We are tasked with verifying whether the constant function y(t) = -4 is a solution to the differential equation y' = t^2(y + 4). To do this, we need to substitute y(t) = -4 into the equation and check if both sides are equal.
2. Step 2: Compute the derivative of y(t). Since y(t) = -4 is a constant function, its derivative y' is 0. This is because the derivative of a constant is always zero.
3. Step 3: Substitute y(t) = -4 and y' = 0 into the differential equation y' = t^2(y + 4). Replace y' with 0 and y with -4 in the equation.
4. Step 4: Simplify the right-hand side of the equation. Substitute y = -4 into the expression t^2(y + 4). This becomes t^2((-4) + 4), which simplifies to t^2(0) = 0.
5. Step 5: Compare both sides of the equation. The left-hand side is y' = 0, and the right-hand side is t^2(y + 4) = 0. Since both sides are equal, y(t) = -4 satisfies the differential equation.