# 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.

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} 