Question

Consider the initial value problem \(y^{\prime}=y^{1 / 3}, y(0)=0\) from Example 3 in the text. (a) Is there a solution that passes through the point \((1,1) ?\) If so, find it. (b) Is there a solution that passes through the point \((2,1)\) ? If so, find it. (c) Consider all possible solutions of the given initial value problem. Determine the set of values that these solutions have at \(t=2\)

Short answer

#Answer# There is no solution passing through the points (1,1) and (2,1). The set of values of the solution at t=2 is { (2/3)^(3/2) }.

Step-by-step solution

Solve the IVP

Given the initial value problem \(y'=y^{1/3}\), with \(y(0)=0\). The given ODE is separable. Separate the variables and integrate both sides: \[ \int \frac{dy}{y^{1/3}}=\int dt \] Now, integrate both sides: \[ 3y^{2/3}=t+C \] Apply the initial condition \(y(0)=0\) to get: \[ 3*0^{2/3}=0+C \Rightarrow C=0 \] Therefore, the solution of the given IVP is: \[ y(t)=\left(\frac{t}{3}\right)^{3/2} \]

Solutions passing through \((1,1)\) and \((2,1)\)

(a) Check if the solution passes through the point \((1,1)\): \[ y(1)=\left(\frac{1}{3}\right)^{3/2}=\frac{1}{3^{3/2}} \] Since \(\frac{1}{3^{3/2}} \neq 1\), there is no solution that passes through the point \((1,1)\). (b) Check if the solution passes through the point \((2,1)\): \[ y(2)=\left(\frac{2}{3}\right)^{3/2} \] Since \(\left(\frac{2}{3}\right)^{3/2} \neq 1\), there is no solution that passes through the point \((2,1)\).

Determine the set of values at \(t=2\)

We have the solution: \[ y(t)=\left(\frac{t}{3}\right)^{3/2} \] Finding the values of \(y\) at \(t=2\): \[ y(2)=\left(\frac{2}{3}\right)^{3/2} \] Hence, the set of values that the solutions have at \(t=2\) is \(\left\{\left(\frac{2}{3}\right)^{3/2}\right\}\).

Key concepts

Separable Differential Equations

Separable differential equations are a special type of differential equations where variables can be easily separated. This means that you can rewrite the equation such that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the other side. For example, consider the original equation in the IVP: \( y' = y^{1/3} \). This can be rearranged to:

• \( \frac{dy}{y^{1/3}} = dt \)

This separation is crucial because it allows us to solve the differential equation by integrating both sides separately. This technique simplifies complex problems and is a fundamental strategy in solving many initial value problems. Being able to identify and manipulate separable differential equations is essential for applying integration techniques effectively.

Integration Techniques

Once we separate the variables in a differential equation, the next step involves integrating both sides of the equation. This is where core integration techniques come into play. In the initial value problem at hand, the integration is relatively straightforward but requires precision:To integrate \( \frac{dy}{y^{1/3}} \), we rewrite it in a more familiar form as \( y^{-1/3} \, dy \). By applying the power rule of integration, we find:

• \( \int y^{-1/3} \, dy = \frac{3}{2}y^{2/3} + C \)

At the same time, the integration of \( dt \) simply gives \( t + C\), where \( C \) is the constant of integration. These integrations help us build the general solution to the original differential equation. Having a solid grasp on these techniques allows for smooth sailing through solving separable differential equations.

Differential Equations Solutions

Solutions to differential equations can vary greatly depending on initial conditions and the form of the differential equation. In our exercise, finding the solution involves utilizing the initial condition \( y(0) = 0 \) to determine any constants in our integrated function. This leads us to the specific solution:

• \( y(t) = \left(\frac{t}{3}\right)^{3/2} \)

This is not just any solution; it is the unique solution that satisfies the given initial condition.Given this specific solution, we then explore whether it can pass through certain points like \((1,1)\) and \((2,1)\). By substituting these points into the solution, we quickly determine if they satisfy the equation. Exploration of these possibilities is essential in understanding the range and limitations of solutions to any initial value problem. The ability to evaluate solutions at specific points is a key skill in interpreting the behaviors modeled by differential equations.