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Find the general solution. \(t y^{\prime}+4 y=0\)

Short Answer

Expert verified
Answer: The general solution to the given differential equation is \(y(t) = \frac{C}{t^4}\), where \(C\) is an arbitrary constant.

Step by step solution

01

Identify Variables and Constants

First, let's identify the variables, constants, and derivatives present in the linear differential equation. Define \(y(t)\) as the dependent variable, then \(y^{\prime}(t)\) as the first derivative of \(y\) with respect to the independent variable \(t\). The problem is defined by the equation: \(t y^{\prime}(t) + 4y(t) = 0\)
02

Rewrite the Equation in Canonical Form

Rewrite the equation in the canonical form for a first-order linear differential equation: \(y^{\prime}(t) + P(t) y(t) = Q(t)\) Dividing both sides of the equation by \(t\) gives us the equation in canonical form: \(y^{\prime}(t) + \frac{4}{t}y(t) = 0\) Comparing the given equation to the general form, we find \(P(t) = \frac{4}{t}\) and \(Q(t) = 0\).
03

Find the Integrating Factor

Calculate the integrating factor \(I(t) = e^{\int{P(t) dt}}\). Integrate \(P(t)\) to find: \(I(t) = e^{\int{\frac{4}{t} dt}}\) \(I(t) = e^{4\ln{t}}\) \(I(t) = t^4\)
04

Multiply the Equation by the Integrating Factor

Multiply both sides of the equation by the integrating factor \(I(t) = t^4\): \(t^4 y^{\prime}(t) + 4t^3 y(t) = 0\)
05

Integrate the Modified Equation

Integrate both sides of the resulting equation: \(\int{(t^4 y^{\prime}(t) + 4t^3 y(t)) dt} = \int{0 dt}\) Recall that if \([I(t) y(t)]^{\prime} = I(t) y^{\prime}(t) + P(t) I(t) y(t)\), then by integrating we have: \(I(t) y(t) = \int{Q(t) I(t) dt} + C\) Here, \(Q(t) = 0\), so the integral vanishes: \(t^4 y(t) = C\)
06

Solve for the General Solution

Finally, divide both sides by the integrating factor to get the general solution of the differential equation: \(y(t) = \frac{C}{t^4}\) The general solution to the given differential equation is \(y(t) = \frac{C}{t^4}\), where \(C\) is an arbitrary constant.

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Key Concepts

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

General Solution of Differential Equation
Understanding the general solution of a differential equation is pivotal as it encompasses all possible solutions to the equation. A differential equation relates a function with its derivatives, representing a range of possible behaviors. For first-order linear differential equations, the general solution can often be expressed in the form of a function plus an arbitrary constant, symbolizing the family of all solutions.

For the equation provided, \(t y'(t) + 4y(t) = 0\), the general solution is expressed as \(y(t) = \frac{C}{t^4}\), where \(C\) is a placeholder for any constant value. This form is derived after performing several steps, which includes isolating the variable and its derivative, finding an integrating factor, and finally, integrating the modified equation. The 'C' represents the infinite number of initial conditions that could apply to specific scenarios, therefore offering a versatile solution that can be tailored to fit various situations.
Integrating Factor
The concept of an integrating factor is a clever technique used to solve first-order linear differential equations. It transforms a non-exact differential equation into an exact one by multiplication, which then allows for straightforward integration. The integrating factor, \(I(t)\), is generally of the form \(e^{\int P(t) dt}\), where \(P(t)\) is a function of the independent variable present in the differential equation.

In our example, \(P(t) = \frac{4}{t}\) yields an integrating factor of \(I(t) = t^4\), after simplifying \(e^{4\ln{t}}\). The key role of the integrating factor is to set up the equation in a way that the left-hand side becomes the derivative of the product of \(I(t)\) and \(y(t)\), which is an essential step towards finding the general solution.
First-Order Linear Differential Equation
A first-order linear differential equation is an equation that involves only the first derivative of the function and the function itself, expressed in the form \(y'(t) + P(t)y(t) = Q(t)\). Here, \(y'(t)\) is the first derivative of \(y\) with respect to \(t\), \(P(t)\) and \(Q(t)\) are functions of \(t\). These types of equations are distinguishable by their linearity in both the function and its derivative.

The equation \(t y'(t) + 4y(t) = 0\) follows this form after dividing by \(t\) to isolate the derivative. The absence of terms like \(y^2(t)\) or \(y'(t)y(t)\) confirms its linearity. By finding an integrating factor and following steps of multiplication and integration, the general solution to this first-order linear equation could be determined, as was shown in the step by step solution provided.
Homogeneous Differential Equation
Homogeneous differential equations are a special category of differential equations where all terms can be expressed as a derivative of the function. Specifically, a first-order, homogeneous linear differential equation has the standard form \(y'(t) + P(t)y(t) = 0\). Notice that the right side of the equation is zero, indicating the absence of a function of the independent variable alone.

In the given equation, \(Q(t)\) is zero, which makes it homogeneous. This homogeneity signifies that the differential equation describes proportional growth or decay processes, and the solutions to these equations will typically involve exponential functions or powers of the independent variable. Additionally, homogeneous differential equations have the advantage of the integral on the right side being zero, which simplifies the solving process, as seen when the integral resulted in constant \(C\).

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Most popular questions from this chapter

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.

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 \(k\).)

A differential equation of the form $$ y^{\prime}=p_{1}(t)+p_{2}(t) y+p_{3}(t) y^{2} $$ is known as a Riccati equation. \({ }^{5}\) Equations of this form arise when we model onedimensional motion with air resistance; see Section 2.9. In general, this equation is not separable. In certain cases, however (such as in Exercises 24-26), the equation does assume a separable form. Solve the given initial value problem and determine the \(t\)-interval of existence. $$ y^{\prime}=t\left(5+4 y+y^{2}\right), \quad y(0)=-3 $$

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ \frac{d y}{d t}=\frac{t}{y}, \quad y(0)=-2 $$

Sometimes a change of variable can be used to convert a differential equation \(y^{\prime}=f(t, y)\) into a separable equation. (a) Consider a differential equation of the form \(y^{\prime}=f(\alpha t+\beta y+\gamma)\), where \(\alpha, \beta\), and \(\gamma\) are constants. Use the change of variable \(z=\alpha t+\beta y+\gamma\) to rewrite the differential equation as a separable equation of the form \(z^{\prime}=g(z)\). List the function \(g(z)\). (b) A differential equation that can be written in the form \(y^{\prime}=f(y / t)\) is called an equidimensional differential equation. Use the change of variable \(z=y / t\) to rewrite the equation as a separable equation of the form \(t z^{\prime}=g(z)\). List the function \(g(z)\).

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