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Chapter 10: Problem 16

Solve the following differential equations: $$ y^{2} y^{\prime}=\tan t $$

Short Answer

Expert verified
The solution is \(y = \sqrt[3]{-3 \ln|\cos t| + C_1}\).

Step by step solution

01

Rewrite the differential equation

Rewrite the given differential equation in a more standard form. The given equation is y^2 y^{\tau} = \tan tRewrite it as:\( y^2 \frac{dy}{dt} = \tan t \)
02

Separate the variables

Separate the variables to isolate terms involving y on one side and terms involving t on the other side. We get:\( y^2 dy = \tan t dt \)
03

Integrate both sides

Integrate both sides with respect to their respective variables:\( \int y^2 dy = \int \tan t dt \)The integrals become:\( \frac{y^3}{3} = - \ln|\cos t| + C \)where C is the integration constant.
04

Solve for y

Solve for y by isolating y on one side of the equation. Multiply both sides by 3 to get: \( y^3 = -3 \ln|\cos t| + C_1 \)Taking the cube root of both sides, we obtain:\( y = \sqrt[3]{-3\ln|\cos t| + C_1} \)

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Key Concepts

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

Separation of Variables
Separation of variables is a powerful method used to solve differential equations. This technique involves rearranging the equation to separate the variables, so each side of the equation contains only one variable (either t or y).
In our example, we have the differential equation: \( y^2 \frac{dy}{dt} = \tan t \)
To separate variables, we rewrite the equation to isolate terms involving y on one side and terms involving t on the other: \( y^2 dy = \tan t dt \)
By doing this, we have successfully separated the variables, which allows us to proceed with integration.
Integration
Integration is a fundamental concept in calculus used to find the antiderivative or the area under a curve. After separating the variables in our differential equation, we need to integrate both sides with respect to their respective variables: \( \frac{d}{dy} \bigg( \frac{y^3}{3} \bigg) = \frac{d}{dt} \bigg( - \ln|\tan| +C \bigg) \)
On the left side of the equation, we integrate y鲁/3 with respect to y. For the right side, we integrate \( \tan t \) with respect to t. These integrations yield the following results: \( \frac{y^3}{3} = - \ln|\tan t | + C \)
Initial Value Problems
An initial value problem (IVP) in differential equations involves finding a specific solution that meets an initial condition, such as y(0)=y鈧�. To solve an IVP, we use the general solution of the differential equation and apply the initial condition to find the value of the integration constant, C.
In our example, if an initial condition is given, ensure you substitute these at the right stage where you solved for C. However, in this case, the example lacks an initial value, so deriving C directly depends on some initial context or conditions.
Solutions of Differential Equations
The final goal of solving a differential equation is to find the function y(t) that satisfies the equation. In our example, we need to isolate y, which means finding y in terms of t.
After integrating and combining both sides of our separated variables: \( \frac{y^3}{3} = - \ln|\tan t | + C_1 \)
Solving for y:
To isolate y, multiply both sides by 3: \( y^3 = -3 \ln|\tan t | + C_1 \).
Finally, take the cube root of both sides: \( y = \frac{1}{3} \bigg( -\frac{ 3}{ \ln |\tan t|} +C_1 \bigg) \)
This y is the solution that satisfies the original differential equation.

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Most popular questions from this chapter

The Los Angeles Zoo plans to transport a California sea lion to the San Diego 'Loo. 'The animal will be wrapped in a wet blanket during the trip. At any time \(t\), the blanket will lose water (due to evaporation) at a rate proportional to the amount \(f(t)\) of water in the blanket, with constant of proportinality \(k=-.3\). Initially, the blanket will contain 2 gallons of seawater. (a) Set up the differential equation satisfied by \(f(t)\). (b) Use Euler's method with \(n=2\) to estimate the amount of moisture in the blanket after 1 hour. (c) Solve the differential equation in part (a) and compute \(f(1)\). (d) Compare the answers in parts (b) and (c) and approximate the error in using Euler's method.

You are given a logistic equation with one or more initial conditions. (a) Determine the carrying capacity and intrinsic rate. (b) Sketch the graph of \(\frac{d N}{d t}\) versus \(N\) in an \(N z\) -plane. (c) In the \(t N\) -plane, plot the constant solutions and place a dashed line where the concavity of certain solutions may change. (d) Sketch the solution curve corresponding to each given initial condition. $$ d N / d t=-.01 N^{2}+N, N(0)=5 $$

Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}+y=e^{-t}+1 $$

A certain drug is administered intravenously to a patient at the continuous rate of \(r\) milligrams per hour. The patient'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.

One or more initial conditions are given for each differential equation in the following exercises. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solutions. Include a \(y z\) -graph if one is not already provided. Always indicate the constant solutions on the \(t y\) -graph whether they are mentioned or not. $$ y^{\prime}=1 / y, y(0)=-1, y(0)=1 $$
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