Question

The solution of the differential equation satisfying initial condition \(y(0)=1\) is given. \(y^{\prime}=1+y^{2} ; \quad y(t)=\tan \left(t+\frac{\pi}{4}\right)\)

Short answer

Question: Verify that the function \(y(t)=\tan \left(t+\frac{\pi}{4}\right)\) is the solution of the differential equation \(y^{\prime}=1+y^{2}\) with the initial condition \(y(0) = 1\). Answer: Yes, \(y(t)=\tan \left(t+\frac{\pi}{4}\right)\) is the solution of the differential equation \(y^{\prime}=1+y^{2}\) with the initial condition \(y(0) = 1\), as both the differential equation and the initial condition are satisfied.

Step-by-step solution

1. Find the derivative of y(t)

To find the derivative of the given function \(y(t)=\tan \left(t+\frac{\pi}{4}\right)\), we can use the chain rule. Recall that the derivative of \(\tan(u)\) with respect to \(u\) is equal to \(\sec^2(u)\). So, the derivative of \(y(t)\) with respect to \(t\) is: \(y^{\prime}(t) = \frac{d}{dt} \left[\tan \left(t+\frac{\pi}{4}\right)\right] = \sec^2 \left(t+\frac{\pi}{4}\right) \cdot \frac{d}{dt} \left(t+\frac{\pi}{4}\right)\) Since the derivative of \(t+\frac{\pi}{4}\) with respect to \(t\) is simply 1, we get: \(y^{\prime}(t) = \sec^2 \left(t+\frac{\pi}{4}\right)\)

2. Substitute y(t) and y'(t) into the differential equation

Now, we will substitute the given function \(y(t)\) and its derivative \(y^{\prime}(t)\) into the differential equation \(y^{\prime}=1+y^{2}\) to verify if it is indeed a solution. Substituting, we get: \(\sec^2 \left(t+\frac{\pi}{4}\right) = 1 + \tan^2 \left(t+\frac{\pi}{4}\right)\) Notice that the left side is the derivative \(y^{\prime}(t)\), and the right side is \(1+y^{2}(t)\). Now, we will use the Pythagorean identity for trigonometric functions, \(\sec^2(u) = 1 + \tan^2(u)\): \(\sec^2 \left(t+\frac{\pi}{4}\right) = 1 + \tan^2 \left(t+\frac{\pi}{4}\right)\) As we can see, both sides are equal, which means the given function satisfies the given differential equation.

3. Verify the initial condition

Finally, we will check if the given function satisfies the initial condition \(y(0) = 1\). We will substitute \(t = 0\) into the function \(y(t)\): \(y(0) = \tan \left(0+ \frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right)\) We know that \(\tan\left(\frac{\pi}{4}\right) = 1\), so the initial condition is: \(y(0) =1\) Since the given function satisfies both the differential equation and the initial condition, we can conclude that it is indeed the solution of the given problem.

Key concepts

Initial Conditions

Initial conditions in the context of differential equations serve as the starting point for finding a particular solution. These are specific values provided for the dependent variable (often represented as 'y') at a particular independent variable (commonly represented as 't' or 'x'). For instance, in the equation provided, the initial condition is expressed as \(y(0)=1\).

When solving a differential equation, the general solution typically consists of a family of curves, each differing by the constants of integration. The initial condition helps us determine the exact value of these constants and hence pinpoints the specific solution curve that corresponds to the problem at hand.

Importance of Verifying Initial Conditions

In verifying a solution, it is not enough to show that the differential equation holds for the proposed function; one must also check that the initial conditions are satisfied. If the solution adheres to the initial conditions, it implies consistency and contextual correctness, anchoring the solution to the specific scenario depicted by the problem.

Trigonometric Identities

Trigonometric identities are critical tools for solving a variety of mathematical problems, including differential equations. They are essentially equations involving trigonometric functions that hold true for all values within their domains.

Common trigonometric identities include the fundamental ratios for sine, cosine, and tangent, as well as larger identities like the angle sum and double angle formulas. These identities allow for the transformation of trigonometric expressions into equivalent forms, which can greatly simplify calculus operations such as differentiation and integration.

Application in Differential Equations

When dealing with differential equations that contain trigonometric functions, known properties and identities can be used to manipulate the equation into a more workable form. In our exercise example, familiarizing yourself with identities like those involving the secant and tangent functions might be particularly advantageous.

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of composite functions. A composite function is one that involves a function within another function, which in notation is written as \( f(g(x)) \).

The chain rule states that to find the derivative of \( f(g(x)) \), you must take the derivative of the outer function evaluated at the inner function and multiply it by the derivative of the inner function. Symbolically, it is expressed as: \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).

Significance in Solving Differential Equations

Understanding and applying the chain rule is critical when solving differential equations involving composite functions. It helps in obtaining the derivative of the solution function correctly, which is essential for verifying if the proposed solution satisfies the differential equation.

Pythagorean Identity

The Pythagorean identity is one of the most widely recognized trigonometric identities, derived from the Pythagorean Theorem of geometry. This identity provides a relationship between the sine and cosine of an angle, but it is most commonly remembered in terms of the secant and tangent, as \( \sec^2(u) = 1 + \tan^2(u) \).

This identity is useful in transforming an expression with a secant function into one with tangent, or vice versa, which is often needed when solving trigonometric equations or simplifying expressions in calculus.

Utilization in Differential Equations

For differential equations that involve trigonometric functions, the Pythagorean identity can transform terms to match the form required in the equation or to verify solutions, as shown in our exercise. This transformation can lead to a simplified process of verification and confirm the validity of a potential solution to a differential equation.