Question

\(2 t+\left(1+y^{2}\right) y^{\prime}=0, \quad y(1)=1\)

Short answer

Question: Solve the first-order differential equation \(2t + (1 + y^2)y' = 0\) with the initial condition \(y(1) = 1\). Answer: The solution to the given first-order differential equation is \(y(t) = \tan\left(-t^2 + \frac{\pi}{4} + 1\right)\).

Step-by-step solution

Rewrite the equation

In this step, we will rewrite the equation by isolating the derivative y'. We start with the given equation: \(2 t+\left(1+y^{2}\right) y^{\prime}=0\) Now, divide by \((1+y^2)\): \(y' = -\frac{2t}{1+y^2}\)

Separate variables

Now, we need to separate the variables, i.e., put all terms related to y on one side of the equation and all terms related to t on the other side. We do this by multiplying both sides by \((1+y^2)\) and dividing both sides by \(-2t\): \((1+y^2)y' = -2t\) \(\frac{dy}{dt} = -\frac{2t}{1+y^2}\) Now we have: \(\frac{dy}{1+y^2} = -\frac{2t}{1} dt\)

Integrate both sides

Next, we integrate both sides of the equation with respect to their respective variables: \(\int \frac{dy}{1+y^2} = -2\int t dt\) The left-hand side can be recognized as the integral of the arctangent function, so we have: \(\arctan(y) = -t^2 + C\)

Apply the initial condition

Now, we apply the initial condition \(y(1) = 1\). Plugging in t = 1 and y = 1 to find the constant C: \(\arctan(1) = -1^2 + C\) \(\frac{\pi}{4} = -1 + C\) \(C = \frac{\pi}{4} + 1\)

Solve for the function y(t)

Finally, we solve for the function y(t) by taking the tangent of both sides of the equation: \(\tan(\arctan(y)) = \tan\left(-t^2 + \frac{\pi}{4}+1\right)\) \(y(t) = \tan\left(-t^2 + \frac{\pi}{4} + 1\right)\) This is the solution to the differential equation with the given initial condition.

Key concepts

Separable Differential Equations

Understanding separable differential equations is crucial for solving many problems in calculus and applied mathematics. These equations can be transformed in such a way that each variable and its differential are on opposite sides of the equation, making them easier to integrate.

Let's take the example from our exercise where we have the differential equation: \(2 t+(1+y^{2}) y'=0, \ y(1)=1\). The first step is to express this in a form that clearly separates the variables t and y. When we rewrite the equation as \(y' = -\frac{2t}{1+y^2}\), we have it in a separable form.Breaking it down further, by cross-multiplying, we arrive at: \(\frac{dy}{1+y^2} = -2t dt\). Here, 'dy' is associated with all terms involving 'y', and 'dt' is associated with terms involving 't', thus achieving an equation where each variable is on one side. This is the essence of separable differential equations: rearranging the equation so that we can apply integration separately to both sides, one with respect to y, and the other with respect to t.

Solving such equations involves two primary steps: separation of variables, as just described, and then integration of both sides, each with respect to its own variable, to obtain the general solution.

Initial Value Problems

In the world of differential equations, initial value problems (IVPs) are a specific type of problem where we are given a differential equation along with specified values for the variable(s) at a particular point. These initial conditions allow us to find a unique, specific solution to a differential equation that would otherwise have infinitely many solutions.

In our given problem, after integrating and finding the general solution, there is still a constant of integration, C, that needs to be determined. IVPs come into play here. We use the initial conditions, which in our case is \(y(1) = 1\), to find C. By substituting '1' for 't' and '1' for 'y' into the general solution, we can solve for the unknown constant.Generally, for an IVP with a differential equation of the form \(dy/dt = f(t, y)\) and an initial condition \(y(t_0) = y_0\), applying the initial condition is essential in finding the exact solution that fits the specific scenario described by the initial values.

Integration Techniques

When solving differential equations, especially separable ones, we often encounter integrals that require specific techniques to solve. Our exercise showcases a classic example where integration techniques must be applied effectively to find a solution.

Here, the integral we need to evaluate is \(\int \frac{dy}{1+y^2} = \arctan(y)\), which is a standard result that can be found in most calculus textbooks. This result comes from knowing the derivatives of inverse trigonometric functions, a key topic in integration techniques. The right side of the equation involves a more straightforward integral, \(-2 \int t dt = -t^2\), which uses the power rule for integration.These techniques are required to handle various forms of integrals, such as integration by parts, trigonometric integrals, partial fractions, and others. Recognizing which technique to use is pivotal and often requires practice to master. For instance, an integral that looks complex could be simplified using trigonometric substitution, or an integral with a polynomial denominator could be broken down into partial fractions for easier integration.