# Many modeling applications of differential equations use the initial value problem (IVP) dy/dt =...

## Question:

Many modeling applications of differential equations use the initial value problem (IVP)

{eq}\frac{dy}{dt} = k(y - A), y(0) = y_0, {/eq}

where k, A, and y_0 are constants. The IVP has the solution

{eq}y(t)=A + (y_0 -A)e^kt. {/eq}

You should memorize this solution and be able to write it down by inspection". The solution makes sense because y(t) = A is an equilibrium solution which is stable if k < 0 and unstable if k > 0.

The case with A = 0 is a very important special case: The IVP

{eq}\frac {dy}{dt} = ky, y(0) = y_0, {/eq}has the solution

{eq}y(t) = y_0 e^kt. {/eq}

Practice by solving the following IVP by inspection:

{eq}\frac{dy}{dt }= -1 y + 1, y(0) = 2, {/eq}This IVP fits the standard pattern with k =

A=

And {eq}y_0 {/eq} =

Therefore, the solution to the IVP is

y(t) =

## Initial Value Problems:

Most differential equations are initial value problems as an initial value is necessary in order for us to get a particular solution for any situation. Some differential equations, like the ones mentioned above, are so common that you will find yourself simply remembering them after long enough.

## Answer and Explanation: 1

Note that we can write the differential equation as

{eq}\begin{align*} \frac{dy}{dx} &= -1y + 1 \\ &= -1 ( y - 1) \end{align*} {/eq}

And so by comparison to the top differential equation form, we see that

{eq}\begin{align*} k &= -1 \\ A &= -1 \end{align*} {/eq}

Since {eq}y_0 = 2 {/eq} a solution is

{eq}\begin{align*} y (t) &= A + (y_0 - A) e^{kt} \\ &= -1 + (2 - (-1) ) e^{-1t} \\ &= 3e^{-t} - 1 \end{align*} {/eq}