Question

Find an initial-value problem whose solution is \(y=\cos x+\int_{0}^{x} e^{-t^{2}} d t\)

Short answer

Question: Given the solution \(y = \cos x + \int_{0}^{x} e^{-t^2} dt\), find the corresponding initial-value problem (IVP). Answer: The IVP can be defined as: Find \(y(x)\) if \(\frac{dy}{dx} = -\sin{x} + e^{-x^2}\) and \(y(0) = 1\).

Step-by-step solution

Differentiate the given function

Differentiate the given function \(y = \cos x + \int_{0}^{x} e^{-t^2} dt\) with respect to x: $$ \frac{dy}{dx} = -\sin{x} + e^{-x^2} $$

Find an initial condition

We will find an initial condition for the given function so that the IVP problem is defined. Let's consider the case when \(x = 0\). Using the given function, find the value of \(y\) when \(x = 0\): $$ y(0) = \cos{(0)} + \int_{0}^{0} e^{-t^2} dt = 1 $$ So we have an initial condition: \(y(0) = 1\).

Write down the initial-value problem

Now that we have the first derivative and an initial condition, we can define the IVP. The initial-value problem is a first-order differential equation with the given initial condition, which can be stated as follows: Find \(y(x)\) if: $$ \frac{dy}{dx} = -\sin{x} + e^{-x^2} $$ and $$ y(0) = 1 $$