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Chapter 9: Problem 20

Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem. $$y^{\prime}(t)=y\left(4 t^{3}+1\right), y(0)=4$$

Short Answer

Expert verified
Question: Determine the solution to the given initial value problem: $$y'(t) = y(4t^3 + 1), y(0)=4$$ Answer: The solution to the given initial value problem is $$y(t) = e^{t^4 + t + \ln(4)}$$.

Step by step solution

01

Determine if the equation is separable

To check if the equation is separable, we rewrite the equation in terms of y and t. The given equation is: $$y'(t) = y(4t^3 + 1)$$ Let's try to rewrite it separating the variables by dividing by y and differentiate with respect to t: $$\frac{dy}{dt} = y(4t^3 + 1)$$ So, we have: $$\frac{dy}{y} = (4t^3+1) dt$$ Here, we have succeeded in separating the variables, which means the equation is separable.
02

Integrate both sides

Now, to solve the initial value problem, we need to integrate both sides of the equation. We do this as follows: $$\int \frac{dy}{y} = \int (4t^3+1) dt$$ On integrating, we get: $$\ln|y| = \int (4t^3+1) dt$$ Now integrate the right-hand side: $$\ln|y| = t^4 + t + C_1$$ To solve for y, we need to use the exponent function: $$y(t) = e^{t^4 + t + C_1}$$
03

Apply the given initial condition

Now, we have to apply the given initial condition, which is \(y(0) = 4\). By plugging in the values, we get: $$4 = e^{0^4 + 0 + C_1}$$ $$4 = e^{C_1}$$ Now, to find the value of \(C_1\), we take the natural logarithm: $$C_1 = \ln(4)$$
04

Write the final answer

We now substitute the value of \(C_1\) back into the equation \(y(t)\) to get the final answer: $$y(t) = e^{t^4 + t + \ln(4)}$$ This is the solution to the given initial value problem.

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Key Concepts

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

Separable Differential Equations
Understanding separable differential equations is key to solving many types of initial value problems. A differential equation is said to be separable if it can be expressed in the form \( \frac{dy}{dx} = g(x)h(y) \), where the function on the right side of the equation is the product of two functions, each dependent on a single variable. In other words, the variables can be separated and then integrated separately.

The process generally involves rearranging the equation to isolate \( dy \) and \( dx \) on different sides. For example, the given equation \( y'(t) = y(4t^3 + 1) \) is transformed into \( \frac{dy}{y} = (4t^3+1) dt \), thereby separating the variables and setting up for integration. It is important to remember that integrals can include an arbitrary constant, which will be determined later by applying the initial conditions.
Integration of Functions
Integration is an essential operation in calculus, often thought of as the reverse process of differentiation, and is used for obtaining the function from its rate of change. In the context of separable differential equations, after separating the variables, you need to integrate both sides to find a general solution.

The integral \( \int \frac{dy}{y} \) represents a common form that you鈥檒l see, resulting in the natural logarithm of the absolute value of \( y \). Similarly, the integration of a function of \( t \) on the right-hand side will vary depending on the form of the function. These integrations may involve simple antiderivatives for polynomials, or more complex integrals for other types of functions.
Applying Initial Conditions
After solving the indefinite integrals on both sides of a separable equation, we get a general solution that includes arbitrary constants. To find the specific solution that satisfies the initial value problem, we apply the initial conditions given in the problem.

In this context, the initial condition \( y(0) = 4 \) provides a specific point on the curve described by the differential equation. By substituting this condition into the general solution, we can solve for the constant \( C_1 \). It is crucial to ensure that the initial condition is applied correctly to find an exact solution that represents the physical or mathematical scenario being modeled.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental constant approximately equal to 2.71828. This function is the inverse of the exponential function and arises naturally in the process of integrating the function \( 1/x \).

In the given exercise, the natural logarithm is used to find the value of the constant \( C_1 \) when the initial condition is applied. With \( \ln(4) \) representing \( C_1 \) in our problem, it allows us to express the solution to the initial value problem with a specific constant, rather than a general one. In calculus and differential equations, understanding the relationship between the exponential function and the natural logarithm is pivotal for solving various types of equations.

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Most popular questions from this chapter

Draining tanks Consider the tank problem in Example 7. For the following parameter values, find the water height function. Then determine the approximate time at which the tank is first empty and graph the solution. $$H=2.25 \mathrm{m}, A=2 \mathrm{m}^{2}, a=0.5 \mathrm{m}^{2}$$

Analyzing models The following models were discussed in Section 9.1 and reappear in later sections of this chapter. In each case carry out the indicated analysis using direction fields. Chemical rate equations Consider the chemical rate equations \(y^{\prime}(t)=-k y(t)\) and \(y^{\prime}(t)=-k y^{2}(t),\) where \(y(t)\) is the concentration of the compound for \(t \geq 0,\) and \(k>0\) is a constant that determines the speed of the reaction. Assume the initial concentration of the compound is \(y(0)=y_{0}>0\). a. Let \(k=0.3\) and make a sketch of the direction fields for both equations. What is the equilibrium solution in both cases? b. According to the direction fields, which reaction approaches its equilibrium solution faster?

Suppose an object with an initial temperature of \(T_{0}>0\) is put in surroundings with an ambient temperature of \(A\) where \(A<\frac{T_{0}}{2} .\) Let \(t_{1 / 2}\) be the time required for the object to cool to \(\frac{T_{0}}{2}\) a. Show that \(t_{1 / 2}=-\frac{1}{k} \ln \left(\frac{T_{0}-2 A}{2\left(T_{0}-A\right)}\right)\) b. Does \(t_{1 / 2}\) increase or decrease as \(k\) increases? Explain. c. Why is the condition \(A<\frac{T_{0}}{2}\) needed?

Consider the differential equation \(y^{\prime \prime}(t)+k^{2} y(t)=0,\) where \(k\) is a positive real number. a. Verify by substitution that when \(k=1\), a solution of the equation is \(y(t)=C_{1} \sin t+C_{2} \cos t .\) You may assume this function is the general solution. b. Verify by substitution that when \(k=2\), the general solution of the equation is \(y(t)=C_{1} \sin 2 t+C_{2} \cos 2 t\) c. Give the general solution of the equation for arbitrary \(k>0\) and verify your conjecture.

A community of hares on an island has a population of 50 when observations begin (at \(t=0\) ). The population is modeled by the initial value problem $$\frac{d P}{d t}=0.08 P\left(1-\frac{P}{200}\right), P(0)=50$$ a. Find and graph the solution of the initial value problem, for \(t \geq 0\) b. What is the steady-state population?
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