Modeling with Differential Equations

Suppose that you bake a potato in your microwave. When you remove the potato from the microwave oven, you record the temperature every two minutes (see this table). If room temperature is 73 degrees, write a differential equation that models the temperature of the potato, $T(t)$, at time $t$.

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Verifying Solutions.

$y(t) = e^{4t}$; $y' = 4y$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Verifying Solutions.

$y(t) = 3e^{-2t}$; $y' = -2y$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Verifying Solutions.

$y(t) = 3e^{5t}$; $y' - 5y = 0$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Verifying Solutions.

$y(t) = e^{3t} - 2$; $y' = 3y + 6$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Verifying Solutions.

$y(t) = -7e^{t^2} - \dfrac{1}{2}$; $y' = 2ty + t$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Verifying Solutions.

$y(t) = (t^8 - t^4)^{1/4}$; $y' = \dfrac{2y^4 + t^4}{ty^3}$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Verifying Solutions.

$y(t) = t$; $y'' - ty' + y = 0$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Verifying Solutions.

$y(t) = e^t + e^{2t}$; $y'' - 4y' + 4y = e^t$

Many differential equations have solutions of the form $y(t) = e^{at}$, where $a$ is some constant. For example, $y(t) = e^{3t}$ is a solution to the equation $y' = 3y$. Find all values of $a$ such that $y(t) = e^{at}$ is a solution to the given equation in Finding Solutions.

$y' = -2y$

Many differential equations have solutions of the form $y(t) = e^{at}$, where $a$ is some constant. For example, $y(t) = e^{3t}$ is a solution to the equation $y' = 3y$. Find all values of $a$ such that $y(t) = e^{at}$ is a solution to the given equation in Finding Solutions.

$y' + 7 y = 0$

Many differential equations have solutions of the form $y(t) = e^{at}$, where $a$ is some constant. For example, $y(t) = e^{3t}$ is a solution to the equation $y' = 3y$. Find all values of $a$ such that $y(t) = e^{at}$ is a solution to the given equation in Finding Solutions.

$y'' - 3y' + 2y = 0$

Many differential equations have solutions of the form $y(t) = e^{at}$, where $a$ is some constant. For example, $y(t) = e^{3t}$ is a solution to the equation $y' = 3y$. Find all values of $a$ such that $y(t) = e^{at}$ is a solution to the given equation in Finding Solutions.

$y'' + 4y' - 12y = 0$

Many differential equations have solutions of the form $y(t) = e^{at}$, where $a$ is some constant. For example, $y(t) = e^{3t}$ is a solution to the equation $y' = 3y$. Find all values of $a$ such that $y(t) = e^{at}$ is a solution to the given equation in Finding Solutions.

$y''' - y'' - 4y' + 4y = 0$

Many differential equations have solutions of the form $y(t) = e^{at}$, where $a$ is some constant. For example, $y(t) = e^{3t}$ is a solution to the equation $y' = 3y$. Find all values of $a$ such that $y(t) = e^{at}$ is a solution to the given equation in Finding Solutions.

$y^{(5)} - 5y''' + 4y' = 0$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Initial Value Problems. Then use the initial conditions to determine the constants $C$ or $c_1$ and $c_2$.

$y' = 4y$, $y(0) = 2$, $y(t) = Ce^{4t}$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Initial Value Problems. Then use the initial conditions to determine the constants $C$ or $c_1$ and $c_2$.

$y' + 7 y = 0$, $y(0) = 2$, $y(t) = C e^{-7t}$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Initial Value Problems. Then use the initial conditions to determine the constants $C$ or $c_1$ and $c_2$.

$y'' + 4y = 0$, $y(0) = 1$, $y'(0) = 0$, $y(t) = c_1 \cos 2t + c_2 \sin 2t$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Initial Value Problems. Then use the initial conditions to determine the constants $C$ or $c_1$ and $c_2$.

$y'' - 5y' + 4y = 0$, $y(0) = 1$, $y'(0) = 0$, $y(t) = c_1 e^t + c_2 e^{4t}$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Initial Value Problems. Then use the initial conditions to determine the constants $C$ or $c_1$ and $c_2$.

$y'' + 4y' + 13y = 0$, $y(0) = 1$, $y'(0) = 0$, $y(t) = c_1 e^{-2t} \cos 3t + c_2 e^{-2t} \sin 3t$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Initial Value Problems. Then use the initial conditions to determine the constants $C$ or $c_1$ and $c_2$.

$y'' - 4y' + 4y = 0$, $y(0) = 1$, $y'(0) = 0$, $y(t) = c_1 e^{2t} + c_2 te^{2t}$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Boundary Value Problems. Then use the boundary conditions to determine the constants $c_1$ and $c_2$ (if possible).

$y'' + 4y = 0$, $y(0) = 1$, $y(\pi) = 0$, $y(t) = c_1 \cos 2t + c_2 \sin 2t$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Boundary Value Problems. Then use the boundary conditions to determine the constants $c_1$ and $c_2$ (if possible).

$y'' - 5y' + 4y = 0$, $y(0) = 1$, $y(1) = 0$, $y(t) = c_1 e^t + c_2 e^{4t}$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Boundary Value Problems. Then use the boundary conditions to determine the constants $c_1$ and $c_2$ (if possible).

$y'' + 4y' + 13y = 0$, $y(0) = 1$, $y(\pi) = 0$, $y(t) = c_1 e^{-2t} \cos 3t + c_2 e^{-2t} \sin 3t$

Use direct substitution to verify that $y(t)$ is a solution of the given differential equation in Boundary Value Problems. Then use the boundary conditions to determine the constants $c_1$ and $c_2$ (if possible).

$y'' - 4y' + 4y = 0$, $y(0) = 1$, $y(1) = 0$, $y(t) = c_1 e^{2t} + c_2 te^{2t}$