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A first-order linear ordinary differential equation, often abbreviated as ODE, is an equation involving the derivatives of a function. In simple terms, it provides information about how a function changes over time. For a first-order ODE, we are particularly interested in equations where only the first derivative is present without higher derivatives. In our example, the equation \( y'(t) = \sin t + \cos 2t \) is first order because it only involves the first derivative \( y' \).

The significance of linearity in an ODE is that it means the function and its derivative are not multiplied together, which can simplify solving the equation. This makes linear ODEs easier to manage and solve as compared to nonlinear ones. For students encountering ODEs, recognizing the order and linearity helps steer toward the right methods of solving them.

Integration is a fundamental mathematical operation that is essentially the reverse of differentiation. When you integrate a derivative, you obtain the original function that was differentiated. For the exercise at hand, integrating the expression \( y'(t) = \sin t + \cos 2t \) involves finding functions whose derivatives are \( \sin t \) and \( \cos 2t \) respectively.

• The integral of \( \sin t \) is \( -\cos t \). This is because differentiating \( -\cos t \) gives back \( \sin t \).
• The integral of \( \cos 2t \) is \( \frac{1}{2}\sin 2t \). The factor of \( \frac{1}{2} \) appears due to the chain rule used in differentiation.

By adding these two integrals together and then including a constant of integration \( C \), we obtain the general solution to the differential equation.

An initial condition is a crucial piece in solving an initial value problem, as it helps determine the specific solution from the general solution that you get after integration. In our problem, the initial condition given is \( y(0) = 4 \).

This means that at the point \( t = 0 \), the value of \( y \) is 4. By substituting this pair into the integrated function, we can solve for the constant of integration \( C \). With this specific value of \( C \), we are able to write the particular solution of the differential equation that satisfies both the equation itself and the initial condition.

The constant of integration, denoted by \( C \), arises because integration returns a family of functions, all differing by a constant. This constant reflects the infinite number of vertical shifts that the family of curves can undergo and still be a valid antiderivative.

When solving an initial value problem, the goal is to decide this constant using the initial condition provided. In our equation, after integrating, we had \( -\cos t + \frac{1}{2}\sin 2t + C \).

• By using the initial condition \( y(0) = 4 \), we substitute 0 into the equation and solve for \( C \).
• This gives us the exact value of \( C = 5 \), which completes the specific solution to the differential equation that satisfies the initial condition.

Finding \( C \) allows us to narrow down to the one solution curve that passes through this initial point, making the solution unique.