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Chapter 10: Problem 7

Solve the following differential equations: $$y^{\prime}=\left(\frac{t}{y}\right)^{2} e^{t^{3}}$$

Short Answer

Expert verified
y = \bigg( \frac{4}{3} e^{t^3} + C \bigg) ^{1/4}

Step by step solution

01

- Write the given differential equation

The given differential equation is: \[ y^{\text{}} = \frac{t}{y}^{2} e^{t^{3}} \]
02

- Separate the variables

Rewrite the equation to separate the variables: \[ y \frac{dy}{dt} = \frac{t^2 e^{t^3}}{y^2} \]Multiply both sides by \(y^2\) and divide by \(t^2 e^{t^3}\) to get: \[ y^3 dy = t^2 e^{t^3} dt \]
03

- Integrate both sides

Integrate both sides of the equation to find the solution:\[ \frac{y^4}{4} = \frac{1}{3} e^{t^3} + C \]where \(C\) is the constant of integration. Rearrange to express \(y\) explicitly:\[ y^4 = \frac{4}{3} e^{t^3} + C \]
04

- Solve for y

Take the fourth root of both sides to solve for \(y\):\[ y = \bigg( \frac{4}{3} e^{t^3} + C \bigg) ^{1/4} \]

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Key Concepts

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

Variable Separation
Variable separation is a technique used to solve differential equations by isolating the variables on different sides of the equation. The aim is to rewrite the equation in a form where all terms involving one variable are on one side and all terms involving the other variable are on the opposite side. In our example, the given differential equation is: Using variable separation, we rewrite the equation as: This effectively separates the variable 'y' on one side and 't' on the other side. This step is important because it sets the stage for integration, allowing us to solve the differential equation step-by-step.
Integration
Once we have separated the variables, the next step is to integrate both sides of the equation. Integration helps us find an antiderivative, which essentially means undoing the differentiation process to get back to the original function. For our example, we get: and Integrating both sides, we find: where is the constant of integration. Integration turns the separated variables into functions that can provide more insight into the solution of the differential equation.
Constant of Integration
The constant of integration, often represented by 'C', is an important concept when integrating. It represents the infinite number of possible antiderivatives for a given function, as each antiderivative differs by a constant value. In our example, after integrating, we have: The constant 'C' accounts for the fact that there could be many solutions to the differential equation, depending on initial or boundary conditions. It鈥檚 essential to include this constant because it enables us to consider all possible solutions, not just a single one.
Solving for Explicit Solution
Finally, after integrating and including the constant of integration, we solve explicitly for the variable of interest, in this case, 'y'. An explicit solution is one where the dependent variable (y) is expressed solely in terms of the independent variable (t). For our equation, we eventually arrive at: Taking the fourth root of both sides, we get: This gives us the explicit solution: Having the explicit solution allows us to easily evaluate 'y' for any given value of 't', providing a clear and direct answer to the original differential equation.

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Most popular questions from this chapter

Radium 226 is a radioactive substance with a decay constant .00043. Suppose that radium 226 is being continuously added to an initially empty container at a constant rate of 3 milligrams per year. Let \(P(t)\) denote the number of grams of radium 226 remaining in the container after years (a) Find an initial-value problem satisfied by \(P(t).\) (b) Solve the initial-value problem for \(P(t).\) (c) What is the limit of the amount of radium 226 in the container as \(t\) tends to infinity?

In economic theory, the following model is used to describe a possible capital investment policy. Let \(f(t)\) represent the total invested capital of a company at time \(t .\) Additional capital is invested whenever \(f(t)\) is below a certain equilibrium value \(E,\) and capital is withdrawn whenever \(f(t)\) exceeds \(E .\) The rate of investment is proportional to the difference between \(f(t)\) and \(E .\) Construct a differential equation whose solution is \(f(t),\) and sketch two or three typical solution curves.

Solve the given equation using an integrating factor. Take \(t>0\). $$6 y^{\prime}+t y=t$$

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A certain drug is administered intravenously to a patient at the continuous rate of \(r\) milligrams per hour. The paticnt's body removes the drug from the bloodstream at a rate proportional to the amount of the drug in the blood, with constant of proportionality \(k=.5\) (a) Write a differential equation that is satisfied by the amount \(f(t)\) of the drug in the blood at time \(t\) (in hours). (b) Find \(f(t)\) assuming that \(f(0)=0 .\) (Give your answer in terms of \(r .)\) (c) In a therapeutic 2 -hour infusion, the amount of drug in the body should reach 1 milligram within 1 hour of administration and stay above this level for another hour. However, to avoid toxicity, the amount of drug in the body should not exceed 2 milligrams at any time. Plot the graph of \(f(t)\) on the interval \(1 \leq t \leq 2,\) as \(r\) varies between 1 and 2 by increments of \(.1 .\) That is, plot \(f(t)\) for \(r=1,1.1,1.2,1.3, \ldots . .2 .\) By looking at the graphs, pick the values of \(r\) that yield a therapeutic and nontoxic 2-hour infusion.
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