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find the solution of the given initial value problem. $$ t^{3} y^{\prime}+4 t^{2} y=e^{-t}, \quad y(-1)=0 $$

Short Answer

Expert verified
Question: Determine the function y(t) that solves the given initial value problem. Differential equation: \(ty^{\prime} + 4y = te^{(-t)}\) Initial condition: \(y(-1) = 0\) Answer: \(y(t) = \frac{-t e^{-t} + e^{-t} - 2e}{t^4}\)

Step by step solution

01

Write the linear differential equation in standard form

To obtain the standard form of the given differential equation, we need to write it as: $$ y^{\prime} + P(t) y = Q(t) $$ In our case, we divide both sides of the given equation by \(t^3\) to isolate \(y^\prime\): $$ y^{\prime}+\frac{4}{t}y=\frac{e^{-t}}{t^3} $$ So, we have \(P(t) = \frac{4}{t}\), and \(Q(t) = \frac{e^{-t}}{t^3}\).
02

Determine the integrating factor

The integrating factor, denoted by \(μ(t)\), is given by: $$ μ(t) = e^{\int P(t) dt} $$ Here, \(P(t) = \frac{4}{t}\). So, we have: $$ μ(t)=e^{\int \frac{4}{t} dt} = e^{4\ln|t|}=t^4 $$
03

Multiply the standard form equation by the integrating factor

We will multiply both sides of the equation \(y^{\prime}+\frac{4}{t}y=\frac{e^{-t}}{t^3}\) by the integrating factor \(t^4\): $$ t^4y^{\prime} + 4t^3 y = t e^{-t} $$ Now, we can notice that the left-hand side of the equation is the derivative of a product: $$ \frac{d}{dt}(t^4y) = t e^{-t} $$
04

Integrate both sides of the equation

Integrate both sides of the equation with respect to t: $$ \int \frac{d}{dt}(t^4y) dt = \int t e^{-t} dt $$ The integration yields: $$ t^4y = \int t e^{-t} dt + C $$ To integrate the right side, we use integration by parts. Let \(u=t\), and \(dv=e^{-t} dt\). Then, \(du = dt\), and \(v=-e^{-t}\). Therefore, $$ \int t e^{-t} dt = -t e^{-t} - \int (-e^{-t}) dt = -t e^{-t} + e^{-t} + k $$ Substituting this back into our equation, we get: $$ t^4y = -t e^{-t} + e^{-t} + k $$ Now we can solve for y(t): $$ y(t) = \frac{-t e^{-t} + e^{-t} + k}{t^4} $$
05

Solve for y(t) using the initial condition

We now substitute the initial condition y(-1) = 0 to find the value of k: $$ 0 = \frac{-(-1) e^{-(-1)} + e^{-(-1)} + k}{(-1)^4} $$ Solving for k, we obtain \(k=-2e\). Finally, we substitute the value of k back into the equation for y(t): $$ y(t) = \frac{-t e^{-t} + e^{-t} - 2e}{t^4} $$ This is the solution for 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.

Linear Differential Equation
Linear differential equations are foundational in differential equations and are defined by their linearity in terms of the function and its derivatives. When we look at an equation like \(t^{3} y^{\text{'} }+4 t^{2} y=e^{-t}\), we are looking at a first-order linear differential equation. Once reformulated to standard form \(y^{\text{'} } + P(t) y = Q(t)\), where P(t) and Q(t) are both functions of \(t\), it becomes clear that the solution involves finding an expression for \(y\) that satisfies this equation for any value of \(t\).
The crucial first step is transforming the given equation into the standard linear form. This structure allows us to utilize specific techniques tailor-made for linear equations, such as finding an integrating factor, which will subsequently enable us to solve the equation in a systematic manner. The beauty of linear differential equations lies in their predictability and the straightforward methods through which we can approach them, thus making these types of problems rewarding for students who enjoy a clear step-by-step process.
Integrating Factor
The integrating factor is a technique used when solving linear differential equations and is a key ally in transforming an unwieldy expression into one ready for integration. The integrating factor, usually denoted as \(\mu(t)\), is derived from an equation's function \(P(t)\).
In the exercise, we find the integrating factor using the formula \(\mu(t) = e^{\int P(t) dt}\), where \(P(t)\) is identified from the standard form of the equation. After calculating, we get \(\mu(t)=t^4\), a power function that, when multiplied with the original differential equation, turns the left-hand side into the derivative of the product of \(\mu(t)\) and the function \(y\). This is where the magic happens: the once separable terms are now linked, facilitating the next step—integration. An integrating factor simplifies the equation in exactly the right way, enabling an elegant path to the solution.
Integration by Parts
Integration by parts is a powerful tool often used when the function we want to integrate is a product of two simpler functions. It derives from the product rule for differentiation and is based on the formula \(\int u dv = uv - \int v du\).
In our problem, when faced with the task of integrating \(t e^{-t}\), we recognize that it is the product of two functions, making integration by parts the ideal choice. By identifying \(u=t\) (which becomes easier to integrate when differentiated) and \(dv=e^{-t} dt\) (which upon integration yields a manageable expression), we allow ourselves to turn a complex integral into a pair of simpler ones. The elegant dance of choosing \(u\) and \(dv\), differentiating, and integrating, then substituting back into our formula, leads us to our desired result—a neater integral that brings us closer to finding our function \(y(t)\). This technique not only serves to solve the integral at hand but also instills an appreciation for the interplay between differentiation and integration.

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Most popular questions from this chapter

Epidemics. The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli in 1760 on smallpox. In more recent years many mathematical models have been proposed and studied for many different diseases. Deal with a few of the simpler models and the conclusions that can be drawn from them. Similar models have also been used to describe the spread of rumors and of consumer products. Suppose that a given population can be divided into two parts: those who have a given disease and can infect others, and those who do not have it but are susceptible. Let \(x\) be the proportion of susceptible individuals and \(y\) the proportion of infectious individuals; then \(x+y=1 .\) Assume that the disease spreads by contact between sick and well members of the population, and that the rate of spread \(d y / d t\) is proportional to the number of such contacts. Further, assume that members of both groups move about freely among each other, so the number of contacts is proportional to the product of \(x\) and \(y .\) since \(x=1-y\) we obtain the initial value problem $$ d y / d t=\alpha y(1-y), \quad y(0)=y_{0} $$ where \(\alpha\) is a positive proportionality factor, and \(y_{0}\) is the initial proportion of infectious individuals. (a) Find the equilibrium points for the differential equation (i) and determine whether each is asymptotically stable, semistable, or unstable. (b) Solve the initial value problem (i) and verify that the conclusions you reached in part (a) are correct. Show that \(y(t) \rightarrow 1\) as \(t \rightarrow \infty,\) which means that ultimately the disease spreads through the entire population.

draw a direction field and plot (or sketch) several solutions of the given differential equation. Describe how solutions appear to behave as \(t\) increases, and how their behavior depends on the initial value \(y_{0}\) when \(t=0\). $$ y^{\prime}=y(3-t y) $$

Consider the initial value problem $$ y^{\prime}=3 t^{2} /\left(3 y^{2}-4\right), \quad y(1)=0 $$ (a) Use the Euler formula ( 6) with \(h=0.1\) to obtain approximate values of the solution at \(t=1.2,1.4,1.6,\) and 1.8 . (b) Repeat part (a) with \(h=0.05\). (c) Compare the results of parts (a) and (b). Note that they are reasonably close for \(t=1.2,\) \(1.4,\) and 1.6 but are quite different for \(t=1.8\). Also note (from the differential equation) that the line tangent to the solution is parallel to the \(y\) -axis when \(y=\pm 2 / \sqrt{3} \cong \pm 1.155 .\) Explain how this might cause such a difference in the calculated values.

A mass of \(0.25 \mathrm{kg}\) is dropped from rest in a medium offering a resistance of \(0.2|v|,\) where \(v\) is measured in \(\mathrm{m} / \mathrm{sec}\). $$ \begin{array}{l}{\text { (a) If the mass is dropped from a height of } 30 \mathrm{m} \text { , find its velocity when it hits the ground. }} \\ {\text { (b) If the mass is to attain a velocity of no more than } 10 \mathrm{m} / \mathrm{sec} \text { , find the maximum height }} \\ {\text { from which it can be dropped. }} \\ {\text { (c) Suppose that the resistive force is } k|v| \text { , where } v \text { is measured in } \mathrm{m} / \mathrm{sec} \text { and } k \text { is a }} \\ {\text { constant. If the mass is dropped from a height of } 30 \mathrm{m} \text { and must hit the ground with a }} \\ {\text { velocity of no more than } 10 \mathrm{m} / \mathrm{sec} \text { , determine the coefficient of resistance } k \text { that is required. }}\end{array} $$

let \(\phi_{0}(t)=0\) and use the method of successive approximations to solve the given initial value problem. (a) Determine \(\phi_{n}(t)\) for an arbitrary value of \(n .\) (b) Plot \(\phi_{n}(t)\) for \(n=1, \ldots, 4\). Observe whether the iterates appear to be converging. (c) Express \(\lim _{n \rightarrow \infty} \phi_{n}(t)=\phi(t)\) in terms of elementary functions; that is, solve the given initial value problem. (d) Plot \(\left|\phi(t)-\phi_{n}(t)\right|\) for \(n=1, \ldots, 4 .\) For each of \(\phi_{1}(t), \ldots . \phi_{4}(t)\) estimate the interval in which it is a reasonably good approximation to the actual solution. $$ y^{\prime}=-y / 2+t, \quad y(0)=0 $$

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