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Chapter 2: Problem 17

The given equation is an implicit solution of \(N(t, y) y^{\prime}+M(t, y)=0\), satisfying the given initial condition. Assuming the equation \(N(t, y) y^{\prime}+M(t, y)=0\) is exact, determine the functions \(M(t, y)\) and \(N(t, y)\), as well as the possible value(s) of \(y_{0}\). \(t^{3} y+e^{t}+y^{2}=5, \quad y(0)=y_{0}\)

Short Answer

Expert verified
Answer: The possible values for the initial condition \(y_0\) are \(2\) and \(-2\).

Step by step solution

01

Check if the given equation is exact

In order to check if the given equation is exact, we first need to rewrite the equation in the form \(N(t, y) y^{\prime}+M(t, y)=0\). Given Equation: \(t^3y+e^t+y^2=5\) Rearrange: \(t^3y+e^t+y^2-5=0\) Now the equation looks like: \(N(t, y)y'+M(t, y)=0\), where \(N(t, y) = t^3\) and \(M(t, y) = e^t+y^2-5\)
02

Compute the partial derivatives of M(t, y) and N(t, y)

Calculate the partial derivatives of \(M(t, y)\) and \(N(t, y)\) with respect to \(t\) and \(y\) respectively to check the exactness of the equation. \(\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^t+y^2-5)\) \(\frac{\partial M}{\partial y} = 2y\) \(\frac{\partial N}{\partial t} = \frac{\partial}{\partial t}(t^3)\) \(\frac{\partial N}{\partial t} = 3t^2\) Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial t}\), the given equation is not exact.
03

Determine the possible values of the initial condition y_0

Since the given equation is not exact, we cannot determine the functions \(M(t,y)\) and \(N(t,y)\). To find the possible value(s) of \(y_0\), we should use the given initial condition \(y(0)=y_0\) and plug it into the original equation. Original Equation: \(t^3y+e^t+y^2=5\) Setting \(t=0\) and \(y=y_0\): \((0)^3y_0+e^0+y_0^2=5\) \(y_0^2 + 1 = 5\) \(y_0^2 = 4\) Now, solving for \(y_0\), we get two possible values: \(y_0 = \pm 2\). The possible values for the initial condition \(y_0\) are \(2\) and \(-2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Implicit Solution
An implicit solution is used when dealing with equations that cannot be easily solved for one variable in terms of another. Such equations are written in a form where both variables are interdependent, like in the given equation \(t^3 y + e^t + y^2 = 5\). Rather than expressing one variable solely in terms of another, an implicit solution keeps them within the equation itself.

In exact differential equations, the given equation \(N(t, y) y^{\prime} + M(t, y) = 0\) starts as an implicit form. We rearrange the original equation to find the functions \(M(t, y)\) and \(N(t, y)\). In our equation, rearranging provides us with \(N(t, y) = t^3\) and \(M(t, y) = e^t + y^2 - 5\). The implication here is that solving for \(y\) directly is complex or maybe not feasible.

An implicit solution is powerful in allowing us to treat both variables equally, fostering deeper relationships between them without needing an explicit solution for one of the variables.
Initial Conditions
Initial conditions are values specified at the start to solve an equation with constraints. In our example, the initial condition is given as \(y(0) = y_0\). This initial value is critical as it anchors the solution to a specific scenario, helping us find unique solutions applicable to real-world circumstances.

We apply the initial condition to find the value of \(y_0\). By inserting \(t = 0\) in the original equation \(t^3 y + e^t + y^2 = 5\), we derive that \(y_0^2 + 1 = 5\), which simplifies to \(y_0^2 = 4\). Solving for \(y_0\) yields two answers: \(y_0 = 2\) and \(y_0 = -2\).

Understanding initial conditions is fundamental because they allow differential equations to have concrete and singular solutions, making them a necessary component for modeling systems from physics to economics.
Partial Derivatives
In calculus, partial derivatives help us understand how a function changes as each variable in it changes, while holding the other variables constant. They are crucial in studying multivariable functions, especially in exact differential equations.

For our problem, we examine functions \(M(t, y)\) and \(N(t, y)\) to verify the exactness of the differential equation. We find the partial derivative of \(M(t, y)\) with respect to \(y\), which results in \(2y\), and the partial derivative of \(N(t, y)\) with respect to \(t\), resulting in \(3t^2\).

Exact differential equations require these partial derivatives to be equal. Since \(2y eq 3t^2\), the initial equation is not exact. While the equation is not exact, the insight gained from comparing these values is essential when considering potential adjustments or solutions.
  • Partial derivatives are the derivative of a multivariable function concerning one variable at a time.
  • Essential for checking exactness in differential equations.
  • Provide insight into how each variable affects the function independently.

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Most popular questions from this chapter

Active oxygen and free radicals are believed to be exacerbating factors in causing cell injury and aging in living tissue. \({ }^{1}\) These molecules also accelerate the deterioration of foods. Researchers are therefore interested in understanding the protective role of natural antioxidants. In the study of one such antioxidant (Hsian-tsao leaf gum), the antioxidation activity of the substance has been found to depend on concentration in the following way: $$ \frac{d A(c)}{d c}=k\left[A^{*}-A(c)\right], \quad A(0)=0 . $$ In this equation, the dependent variable \(A\) is a quantitative measure of antioxidant activity at concentration \(c\). The constant \(A^{*}\) represents a limiting or equilibrium value of this activity, and \(k\) is a positive rate constant. (a) Let \(B(c)=A(c)-A^{*}\) and reformulate the given initial value problem in terms of this new dependent variable, \(B\). (b) Solve the new initial value problem for \(B(c)\) and then determine the quantity of interest, \(A(c)\). Does the activity \(A(c)\) ever exceed the value \(A^{*}\) ? (c) Determine the concentration at which \(95 \%\) of the limiting antioxidation activity is achieved. (Your answer is a function of the rate constant \(k\).)

Consider the initial value problem $$ \frac{d P}{d t}=r(t)\left(1-\frac{P}{P_{e}}\right) P, \quad P(0)=P_{0} . $$ Observe that the differential equation is separable. Let \(R(t)=\int_{0}^{t} r(s) d s\). Solve the initial value problem. Note that your solution will involve the function \(R(t)\).

State an initial value problem, with initial condition imposed at \(t_{0}=0\), having implicit solution \(y e^{y}+t^{2}=\sin t\).

On August 24,1894 , Pop Shriver of the Chicago White Stockings caught a baseball dropped (by Clark Griffith) from the top of the Washington Monument. The Washington Monument is \(555 \mathrm{ft}\) tall and a baseball weighs \(5 \frac{1}{8} \mathrm{oz}\). (a) If we ignore air resistance and assume the baseball was acted upon only by gravity, how fast would the baseball have been traveling when it was \(7 \mathrm{ft}\) above the ground? (b) Suppose we now include air resistance in our model, assuming that the drag force is proportional to velocity with a drag coefficient \(k=0.0018 \mathrm{lb}-\mathrm{sec} / \mathrm{ft}\). How fast is the baseball traveling in this case when it is \(7 \mathrm{ft}\) above the ground?

Oscillating Flow Rate A tank initially contains \(10 \mathrm{lb}\) of solvent in \(200 \mathrm{gal}\) of water. At time \(t=0\), a pulsating or oscillating flow begins. To model this flow, we assume that the input and output flow rates are both equal to \(3+\sin t \mathrm{gal} / \mathrm{min}\). Thus, the flow rate oscillates between a maximum of \(4 \mathrm{gal} / \mathrm{min}\) and a \(\mathrm{minimum}\) of \(2 \mathrm{gal} / \mathrm{min}\); it repeats its pattern every \(2 \pi \approx 6.28 \mathrm{~min}\). Assume that the inflow concentration remains constant at \(0.5 \mathrm{lb}\) of solvent per gallon. (a) Does the amount of solution in the tank, \(V\), remain constant or not? Explain. (b) Let \(Q(t)\) denote the amount of solvent (in pounds) in the tank at time \(t\) (in minutes). Explain, on the basis of physical reasoning, whether you expect the amount of solvent in the tank to approach an equilibrium value or not. In other words, do you expect \(\lim _{t \rightarrow \infty} Q(t)\) to exist and, if so, what is this limit? (c) Formulate the initial value problem to be solved. (d) Solve the initial value problem. Determine \(\lim _{t \rightarrow \infty} Q(t)\) if it exists.
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