Let's analyze the given differential equation:

$\frac{dT}{dt} = 5(T - 20)$

1. Test $T(t) = 20$:

The derivative of $T(t) = 20$ with respect to $t$ is:

$\frac{dT}{dt} = 0$

Now, substitute $T = 20$ into the right-hand side of the equation:

$5(T - 20) = 5(20 - 20) = 0$

Both sides are equal to zero, so $T(t) = 20$ satisfies the differential equation.

2. Test $T(t) = 20e^{5t} - 20$:

First, differentiate $T(t) = 20e^{5t} - 20$:

$\frac{dT}{dt} = 20 \cdot 5e^{5t} = 100e^{5t}$

Now, substitute $T(t) = 20e^{5t} - 20$ into the right-hand side of the differential equation:

$5(T - 20) = 5(20e^{5t} - 20 - 20) = 5(20e^{5t} - 40) = 100e^{5t} - 200$

Clearly, $\frac{dT}{dt} = 100e^{5t}$, which does not match $100e^{5t} - 200$, so $T(t) = 20e^{5t} - 20$ does not satisfy the equation.

3. Test $T(t) = e^{5t} + 20$:

Differentiate $T(t) = e^{5t} + 20$:

$\frac{dT}{dt} = 5e^{5t}$

Now, substitute $T(t) = e^{5t} + 20$ into the right-hand side:

$5(T - 20) = 5(e^{5t} + 20 - 20) = 5e^{5t}$

Since both sides are equal, $T(t) = e^{5t} + 20$ satisfies the differential equation.

4. Test $T(t) = 20e^{5t} + 20$:

Differentiate $T(t) = 20e^{5t} + 20$:

$\frac{dT}{dt} = 100e^{5t}$

Substitute $T(t) = 20e^{5t} + 20$ into the right-hand side:

$5(T - 20) = 5(20e^{5t} + 20 - 20) = 100e^{5t}$

Since both sides are equal, $T(t) = 20e^{5t} + 20$ satisfies the differential equation.

Conclusion:

The functions that satisfy the differential equation are:

• $T(t) = 20$
• $T(t) = e^{5t} + 20$
• $T(t) = 20e^{5t} + 20$

Would you like further details or have any questions on this solution?

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Related Questions:

1. How do we solve a first-order linear differential equation like $\frac{dT}{dt} = 5(T - 20)$?
2. What is the general form of the solution for a first-order linear differential equation?
3. How does the solution $T(t) = 20e^{5t} - 20$ fail to satisfy the differential equation?
4. Can an exponential function represent the steady-state behavior of this differential equation?
5. What are the key characteristics of solutions to differential equations of this type?

Tip:

When solving differential equations, it's crucial to compute the derivative of each function and substitute it into the equation to check if both sides match.