Question

In Exercises \(21-24\) , solve the initial value problem using the Fundamental Theorem. (Your answer will contain a definite integral.) \(\frac{d y}{d x}=\sin \left(x^{2}\right)\) and \(y=5\) when \(x=1\)

Short answer

The solution to the initial value problem \(\frac{dy}{dx} = \sin(x^2)\), \(y(1) = 5\) is \(y = \int_{1}^{x} \sin(t^{2}) dt + 5\).

Step-by-step solution

Integration

Perform the integration: \[ Y(x) = \int sin(x^{2}) dx\]. This cannot be integrated in terms of elementary functions, but it does not pose an issue here. We will end up with a an arbitrary constant of integration, which we will determine using the given initial conditions.

Applying the Initial Condition

The general solution of this integral is \(y = \int \sin(x^{2}) dx + C\). We substitute the initial condition \(x = 1\) and \(y = 5\) into that formula: 5 = \(\int_{1}^{1} sin(t^{2}) dt + C\). Given that the integral from a number to the same number is zero, we find that the constant \(C = 5\).

Final Solution

Substitute the value of \(C\) into the general solution to obtain the particular solution for our initial value problem: \(y = \int_{1}^{x} \sin(t^{2}) dt + 5\). This is the unique solution of the differential equation which satisfies the given initial condition.