Initial Value Problems

# Initial Value Problems

When we solve differential equations, often times we will obtain many if not infinitely many solutions. For example, consider the differential equation $\frac{dy}{dx} = y$. All solutions to this differential equation are given as $y = Ce^x$ where $C$ is a constant. We can verify this because $\frac{d}{dx} \left ( Ce^x \right ) = Ce^x$. However, suppose that instead we wanted to find a specific solution to our differential equation.

For example, suppose that we look at $\frac{dy}{dx} = y$ again, and suppose that we also want $y$ such that $y(0) = 3$. Since our solution set is $y = Ce^x$, we see that $3 = y(0) = Ce^0 = C$ and so $C = 3$. Therefore, the solution $y = 3e^x$ both satisfies $\frac{dy}{dx} = y$ and $y(0) = 3$. This is what we essentially call an initial value problem where $y(0) = 3$ is the initial value.

 Definition: An Initial Value Problem (often times abbreviated I.V.P.) is a problem where we want to find a solution to some differential equation that satisfies a given Initial Value $y(x_0) = y_0$.

As we learn various techniques of solving differential equations, we will often include solving initial value problems in the process.

## Example 1

Suppose that the solutions to $\frac{dy}{dt} +2yt = 2te^{-t^2}$ are $y = t^2 e^{-t^2} + Ce^{-t^2}$. Find the solution that satisfies both the differential equation and the initial value $y(0) = 5$, and the solution that satisfies both the differential equation and the initial value $y(1) = 4$.

We note that if $y(0) = 5$ then:

(1)
\begin{align} \quad 5 = 0^2 e^{-0^2} + Ce^{-0} \\ \quad 5 = 0 + C \\ \quad C = 5 \end{align}

Therefore $y = t^2 e^{-t^2} + 5e^{-t^2}$ satisfies the initial value $y(0) = 5$.

Now if $y(1) = 4$ then:

(2)
\begin{align} \quad 4 =e^{-1} + Ce^{-1} \\ \quad 4 = \frac{1}{e} + \frac{C}{e} \\ \quad 4e = 1 + C \\ \quad C = 4e - 1 \end{align}

Therefore $y = t^2 e^{-t^2} + (4e - 1)e^{-t^2}$ satisfies the initial value $y(1) = 4$.