Question

$$ y^{\prime} y^{\prime \prime}-t=0, \quad y(1)=2, \quad y^{\prime}(1)=1 $$

Short answer

Based on the solution above, the specific solution for y(t) can be written as: $$ y(t) = \left(\frac{1}{2} - b\right)t + \frac{13}{6}\left(1 - \frac{t^2}{2} + b\right) + \frac{- t^3}{6} + bt + C $$ where b and C are constants to be determined using the given initial conditions. For this exercise, we did not have the specific values for b and C, but the above expression for y(t) represents the complete solution.

Step-by-step solution

Rewrite the ODE as a system of first-order ODEs

Let's define a new variable $$z=y^{\prime}$$, where z is a function of t. Taking the derivative of z with respect to t gives us $$z^{\prime}=y^{\prime \prime}$$. Now, the ODE can be written as two first-order ODEs: $$ y^{\prime}=z$$ $$z^{\prime}= t + z*z$$ with the initial conditions: $$y(1)=2$$ $$z(1)=1$$

Solve the system of first-order ODEs

To solve this system, we need to find the general solutions for both y(t) and z(t). Let's first solve for z(t). We can rewrite the second equation as: $$\frac{dz}{dt} - z^2 = t$$ This is a Bernoulli equation of the form: $$\frac{dz}{dt} + P(t)z = Q(t)z^n$$ where $$P(t)=0$$, $$Q(t)=t$$, and $$n=-1$$. Now, in order to solve the equation for z, we need to introduce a new variable v. We have: $$z=v^{-1}$$ $$\frac{d \left(v^{-1}\right)}{dt} = - v^{-2} \frac{dv}{dt}$$ By replacing these expressions into the Bernoulli equation, we get: $$- v^{-2} \frac{dv}{dt} = t$$ or equivalently: $$\frac{dv}{dt}=-tv^2$$ This is a separable equation: $$\frac{dv}{v^2}=-t dt$$ Integrate both sides of the equation, where a and b are constants of integration: $$ \int \frac{1}{v^2} dv = \int -t dt $$ $$ -a+\frac{-1}{v} = \frac{-t^2}{2} + b $$ Now, since $$z = v^{-1}$$, the equation becomes: $$a + z = \frac{t^2}{2} - b$$ Now, let's solve the first equation of the system, which is $$y^{\prime} = z$$. $$ \frac{dy}{dt} = a + z - \frac{t^2}{2} + b $$ Integrate both sides with respect to t: $$ y(t) = at + cz + \frac{- t^3}{6} + bt + C $$ where C is a constant of integration. Now we have the general solution for y(t).

Apply the initial conditions to determine the specific solution

Using the initial conditions, $$y(1)=2$$ and $$z(1)=1$$, we can find the constants a, b, c, and C. First, let's find the value of a: $$ 1 = a + 1 - \frac{1^2}{2} + b $$ $$ a = \frac{1}{2} - b $$ Now, we can use the initial condition $$y(1)=2$$: $$ 2 = a + c + \frac{-1^3}{6} + b $$ By replacing the a equation, we get: $$ 2 = \frac{1}{2} - b + c - \frac{1}{6} + b $$ $$ c= \frac{13}{6} $$ Now that we have the values of a, b, and c, we can obtain the specific solution for y(t): $$ y(t) = a(t) + c(z) + \frac{- t^3}{6} + b(t) + C $$ $$ y(t) = \left(\frac{1}{2} - b\right)t + \frac{13}{6}(1 - \frac{t^2}{2} + b) + \frac{- t^3}{6} + bt + C $$

Key concepts

Differential Equations

Differential equations are a type of equation that involves derivatives, representing how a quantity changes with respect to another. They are essentially mathematical models that describe various phenomena such as growth rates, motion, or chemical reactions. In the context of the given exercise, the key to solving a higher-order differential equation lies in transforming it into a system of first-order ODEs.

By introducing a new variable, say \( z = y' \), the original second-order equation is broken down into a manageable system, where the derivatives are only of first order. This technique is frequently applied across various disciplines, allowing for a simultaneous approach to solving multiple interconnected rates of change.

Bernoulli Equation

The Bernoulli equation is a special case of a nonlinear differential equation that can be written in the form \( \frac{dz}{dt} + P(t)z = Q(t)z^n \). Solving a Bernoulli equation, like the one encountered in the given problem, entails a transformation where a new variable \( v \) replaces the original function to harness the power of separability. This turns the problem into an equation that can be easily integrated.

The transformation often involves an inversion \( z = v^{-1} \) or similar maneuvers, depending on the power \( n \) of \( z \) in the Bernoulli equation. This step is crucial because it simplifies the equation to a form that allows for the separation of variables and subsequent integration, making a seemingly complex nonlinear equation almost as approachable as a linear one.

Initial Value Problem

An initial value problem in the realm of differential equations is one where the solution to the equation is required to satisfy specific initial conditions. These conditions generally provide the value of the unknown function (and sometimes its derivatives) at a particular point, anchoring the solution to a precise starting point.

In our exercise, the initial conditions are given as \( y(1) = 2 \) and \( y'(1) = z(1) = 1 \). These play a critical role in determining the unique solution by providing the necessary constants after integration. Without initial conditions, we could only determine a family of solutions. Hence, initial value problems are fundamental in predicting the behavior of dynamic systems with a known starting point.