Question

In each of Problems 40 through 43 solve the given initial value problem using the methods of Problems 28 through \(39 .\) $$ y^{\prime} y^{\prime \prime}=2, \quad y(0)=1, \quad y^{\prime}(0)=2 $$

Short answer

Question: Find the function y(t) that satisfies the given second-order, nonlinear ordinary differential equation (ODE) and initial conditions: $$ y^{\prime} y^{\prime \prime} = 2, \quad y(0) = 1, \quad y^{\prime}(0) = 2 $$ Answer: The function y(t) that satisfies the given ODE and initial conditions is: $$ y(t) = \frac{1}{6}(4t + 4)^{3/2} + \frac{2}{3} $$

Step-by-step solution

Integrate the given ODE once with respect to t

Start by multiplying both sides of the equation by dt, and then integrate: $$ \int y^{\prime} y^{\prime \prime} dt = \int 2 dt \\ \int y^{\prime} d(y^{\prime}) = \int 2 dt $$ Now integrate both sides: $$ \frac{1}{2}(y^{\prime})^2 = 2t + C_1 $$

Apply the initial condition y'(0)

Use the initial condition y'(0) = 2: $$ \frac{1}{2}(2)^2 = 2(0) + C_1 \\ \Rightarrow C_1 = 2 $$ So, $$ \frac{1}{2}(y^{\prime})^2 = 2t + 2 $$

Integrate again to find y(t)

To find y(t), we must integrate the expression for y'(t). First, rearrange the equation to make it easier to integrate: $$ y^{\prime} = \sqrt{4t + 4} \\ $$ Now integrate with respect to t: $$ \int y^{\prime} dt = \int \sqrt{4t + 4} dt \\ y(t) = \int \sqrt{4t + 4} dt + C_2 $$ To find the antiderivative, use substitution. Let \(u = 4t + 4\). Then, du = 4dt. Rewrite the integral as: $$ y(t) = \frac{1}{4} \int \sqrt{u} du + C_2 \\ = \frac{1}{4} \cdot \frac{2}{3} u^{3/2} + C_2 \\ = \frac{1}{6} (4t + 4)^{3/2} + C_2 $$

Apply the initial condition y(0)

Use the initial condition y(0) = 1: $$ 1 = \frac{1}{6}(4(0) + 4)^{3/2} + C_2 \\ \Rightarrow C_2 = \frac{2}{3} $$ So the solution to the initial value problem is: $$ y(t) = \frac{1}{6}(4t + 4)^{3/2} + \frac{2}{3} $$

Key concepts

Differential Equations

A differential equation is an equation involving a function and its derivatives. It's like a mathematical recipe that tells you how a quantity changes over time. In other words, a differential equation describes a relationship between a function and the rates at which it changes. Here, we dealt with a second-order differential equation, which involves the second derivative of the function. Understanding these equations is crucial because they model many real-world phenomena such as population growth, heat transfer, or fluid dynamics. In our particular problem, we encountered the equation \( y^{\prime} y^{\prime\prime} = 2 \), which means the product of the first and second derivatives of \( y(t) \) equals 2. This particular form, though unusual, still demands finding a function \( y(t) \) that satisfies this relationship, given certain conditions.

Integration Techniques

Integration techniques are methods used to find antiderivatives or integrals of functions. In this problem, we applied integration twice to solve the given differential equation. At each step, different techniques can be utilized based on the form of the equation. First, we treated \( y^{\prime} \) as a variable, setting up an integral involving \( y^{\prime} \) on the left and simply integrating \( 2 \) on the right. This gives us an intermediate result: \( \frac{1}{2}(y^{\prime})^2 = 2t + C_1 \). After applying the initial condition, this equation was used in the second round of integration. Substitution is often a handy tool in integration, particularly when dealing with square roots or polynomials, as seen when transforming \( \sqrt{4t + 4} \) by introducing a substitution \( u = 4t + 4 \). Such techniques simplify the integration process, allowing us to resolve complex functions into more manageable forms.

Initial Conditions

Initial conditions are specific values provided for the function and its derivatives at a certain point, often used to find particular solutions to differential equations. They are crucial because a differential equation alone may have infinitely many solutions; initial conditions help us narrow this down to a specific solution that satisfies the given criteria. In our problem, we were given the initial conditions \( y(0) = 1 \) and \( y^{\prime}(0) = 2 \). These values are used to determine the constants \( C_1 \) and \( C_2 \) in our integrated equations. The first initial condition helped us establish \( C_1 \) by setting \( t = 0 \) in the equation for \( y^{\prime}(t) \), resulting in \( C_1 = 2 \). Similarly, \( y(0) = 1 \) was essential to solve for \( C_2 \), thus providing a clear and specific solution.

Antiderivatives

Antiderivatives, or indefinite integrals, are the reverse process of differentiation. They allow us to go from a derivative back to the original function. In this exercise, finding the antiderivative of \( y^{\prime} \) was necessary to determine \( y(t) \). Our task started with integrating \( y^{\prime} \) which led to an expression involving \( y^{\prime\prime} \). The first antiderivative found was \( \frac{1}{2}(y^{\prime})^2 = 2t + 2 \), which helped us relate \( y^{\prime} \) back to \( t \). Subsequently, the equation for \( y^{\prime} = \sqrt{4t + 4} \) required determining its antiderivative to ultimately resolve \( y(t) \). The process involved variable substitution to simplify integration, ensuring the final expression was manageable, allowing us to integrate with respect to \( t \) effectively.