Question

Solving initial value problems Solve the following initial value problems. $$y^{\prime }(t)=\sin t+\cos 2t,y(0)=4$$

Short answer

Based on the solution above, determine the value of y(t) at t = π/2. Using the expression for y(t), we substitute t = π/2: $$y(\frac{\pi }{2})=-\cos (\frac{\pi }{2})+\frac{1}{2}\sin (2\cdot \frac{\pi }{2})+5$$ From trigonometry, we know that $$\cos (\frac{\pi }{2})=0$$ and $$\sin (\pi )=0$$. Thus, we have: $$y(\frac{\pi }{2})=-0+\frac{1}{2}\cdot 0+5$$ $$y(\frac{\pi }{2})=5$$

Step-by-step solution

Integrate the equation to find y(t)

First, let's find the function y(t) by integrating both sides of the equation with respect to t. $$\int y^{\prime }(t)dt=\int (\sin t+\cos 2t)dt$$The integral of the left side is simply the function y(t). On the right side, we have the integrals of the sine and cosine functions: $$\int \sin tdt+\int \cos 2tdt$$. Now we need to integrate these functions. To integrate $$\sin t$$, we have $$\int \sin tdt=-\cos t$$ Next, integrating $$\cos 2t$$ involves a substitution. Let $$u=2t$$, so $$du/dt=2$$. Now, we have $$\int \cos 2tdt=\frac{1}{2}\int \cos udu$$ Integrating $$\cos u$$, we have $$\frac{1}{2}\int \cos udu=\frac{1}{2}\sin u$$ Now, substitute back for u, we get $$\frac{1}{2}\sin (2t)$$ Adding the results of the integrations, we get $$y(t)=-\cos t+\frac{1}{2}\sin (2t)+C$$ where C is the constant of integration.

Use the initial condition to find the unknown constant of integration

Now we need to find the value of C using the initial condition $$y(0)=4$$. Substitute t = 0 into the equation for y(t), we get $$y(0)=-\cos (0)+\frac{1}{2}\sin 0+C$$ From trigonometry, we know that $$\cos (0)=1$$ and $$\sin (0)=0$$. So, we have $$4=-1+0+C$$ Solving for C, we get $$C=5$$

Combine the results to obtain the solution of the initial value problem

Now we have all the pieces to find the solution to the initial value problem. The final solution for y(t) is $$y(t)=-\cos t+\frac{1}{2}\sin (2t)+5$$

Key concepts

Integrating Trigonometric Functions

Integrating trigonometric functions is a fundamental skill when solving differential equations involving terms like $\sin t$ and $\cos t$.

While integrating, the key is to remember the basic integrals: $\int \sin tdt=-\cos t$ and $\int \cos tdt=\sin t$. However, when the trigonometric function has a coefficient within the argument, such as $\cos 2t$, we need to use a substitution method for integration.

For example, let's consider the term $\cos 2t$. We set $u=2t$ which leads to $du=2dt$, and thus $dt=\frac{1}{2}du$. Substituting into the integral, we get $\frac{1}{2}\int \cos udu$, which simplifies to $\frac{1}{2}\sin u$. When we replace $u$ with $2t$, we obtain $\frac{1}{2}\sin (2t)$.

This method allows us to break down more complex integral expressions into simpler ones that can be easily integrated using basic knowledge of trigonometry.

Constant of Integration

When we integrate any function, we introduce an arbitrary constant, known as the constant of integration, denoted by $C$. This constant represents the infinite number of antiderivatives a function can have, all of which differ by a constant value.

For instance, after integrating $\int (\sin t+\cos 2t)dt$, we obtain $y(t)=-\cos t+\frac{1}{2}\sin (2t)+C$. The $C$ captures all possible vertical shifts of the antiderivative's graph. To determine its value, we use additional information, often called an initial condition.

In the given exercise, we use the fact that $y(0)=4$ to find the specific value of $C$. When $t=0$, the equation simplifies using basic trigonometric identities, allowing us to solve for $C$ uniquely. By substituting the known initial values into the integrated function, we can calculate the exact value of this constant, making our general solution a particular one that satisfies all given conditions.

Solving Differential Equations

Solving differential equations often involves finding a function that satisfies a given relation between its derivatives. In the case of simple first-order differential equations, we integrate both sides with respect to the independent variable, which is typically time or space.

The process involves recognizing the type of differential equation and then applying the appropriate method – in this case, direct integration. After the integration, we obtain a general solution that includes a constant of integration.

However, in order to determine a unique solution for a differential equation, an initial value or boundary condition is required. This is known as an initial value problem. By applying the given condition, as we did in Step 2 of our example, we can determine the precise value for the constant of integration, leading to the specific solution that satisfies both the differential equation and the initial condition: $y(t)=-\cos t+\frac{1}{2}\sin (2t)+5$.

This process ties together the need to not only integrate trigonometric functions but also to precisely understand the significance of the constant of integration to wholly solve the differential equation at hand.