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Solve the differential equation by the method of integrating factors. \(2 \frac{d y}{d x}+4 y=1\)

Short Answer

Expert verified
The solution to the differential equation is \( y = \frac{1}{4} + C e^{-2x} \).

Step by step solution

01

Write the Equation in Standard Form

Begin by rewriting the differential equation in the standard linear form, which is \[ \frac{d y}{d x} + P(x) y = Q(x). \]Divide the entire equation by 2 to isolate \( \frac{d y}{d x} \).\[ \frac{d y}{d x} + 2y = \frac{1}{2}. \]
02

Identify the Integrating Factor

The integrating factor \( \mu(x) \) for a first-order linear differential equation is found using\[ \mu(x) = e^{\int P(x) \, dx}. \]Here, \( P(x) = 2 \), so we calculate\[ \mu(x) = e^{\int 2 \, dx} = e^{2x}. \]
03

Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor\[ e^{2x}. \]This gives\[ e^{2x} \frac{d y}{d x} + 2e^{2x} y = \frac{1}{2} e^{2x}. \]
04

Rewrite the Left Side as a Derivative

Observe that the left-hand side is the derivative of the product \( e^{2x} y \). Thus,\[ \frac{d}{dx} (e^{2x} y) = \frac{1}{2} e^{2x}. \]
05

Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx} (e^{2x} y) \, dx = \int \frac{1}{2} e^{2x} \, dx. \]The left side integrates to \( e^{2x} y \), and the right side becomes\[ \frac{1}{4} e^{2x} + C, \]where \( C \) is the constant of integration.
06

Solve for y

We now have\[ e^{2x} y = \frac{1}{4} e^{2x} + C. \]Divide both sides by \( e^{2x} \) to solve for \( y \):\[ y = \frac{1}{4} + C e^{-2x}. \]

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Key Concepts

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

Integrating Factors
An integrating factor is a crucial tool used in solving linear first-order differential equations. Its main purpose is to transform a non-solvable differential equation into one that is easy to integrate. The integrating factor is often symbolized as \(\mu(x)\). To find \(\mu(x)\), you calculate the exponential of the integral of the function \(P(x)\) from the standard form of the differential equation. For example, in the equation \( \frac{dy}{dx} + P(x)y = Q(x) \), if \(P(x) = 2\), then the integrating factor becomes:
  • \(\mu(x) = e^{\int 2 \, dx} = e^{2x}\).
Once you have determined \(\mu(x)\), you multiply the entire differential equation by it, simplifying the left side into a single derivative expression, which can then be easily integrated.
Linear Differential Equations
Linear differential equations are equations that can be expressed in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). These are particularly important in calculus and engineering because they describe a wide variety of phenomena, from electric circuits to population dynamics. When a differential equation is linear, superposition applies, which means that the principle of additivity and homogeneity is valid. This makes them simpler to handle compared to non-linear equations.To solve a linear differential equation:
  • First, express it in the standard form.
  • Then, find the integrating factor \(\mu(x)\).
  • Use \(\mu(x)\) to eliminate the \(dy/dx\) term making integration straightforward.
First-Order Differential Equations
First-order differential equations refer to equations where the highest derivative is the first derivative, i.e., \( \frac{dy}{dx} \). These equations are common in many fields such as physics, biology, and economics, often describing processes where the rate of change depends on the initial position or concentration. A typical form for these equations is \( \frac{dy}{dx} = f(x, y) \), emphasizing the function can be dependent on both \(x\) and \(y\).A first-order linear differential equation might look like:
  • \( \frac{dy}{dx} + P(x)y = Q(x) \)
These can be solved using techniques like separation of variables, exact equations, and especially integrating factors.
Solution Technique
The key to solving first-order linear differential equations using integrating factors involves a structured approach:
  • Begin by writing the equation in the standard form: \( \frac{dy}{dx} + P(x)y = Q(x) \).
  • Identify \(P(x)\) in the equation and calculate the integrating factor \(\mu(x) = e^{\int P(x) \, dx}\).
  • Multiply the entire equation by \(\mu(x)\) to form a single derivative on the left side.
  • Integrate both sides to solve for \(y\).
  • Solve for the constant using initial conditions if provided, resulting in the complete solution.
Following these steps not only simplifies the process but ensures that even complex-seeming problems are broken down into manageable parts.

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Most popular questions from this chapter

Solve the differential equation by the method of integrating factors. \(\frac{d y}{d x}+2 x y=x\)

A rocket, fired upward from rest at time \(t=0 .\) has an initial mass of \(m_{0}\) (including its fuel). Assuming that the fuel is consumed at a constant rate \(k\), the mass \(m\) of the rocket, while fuel is being burned, will be given by \(m=m_{0}-k t .\) It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed \(c\) relative to the rocket, then the velocity \(v\) of the rocket will satisfy the equation$$ m \frac{d v}{d t}=c k-m g $$ where \(g\) is the acceleration due to gravity. (a) Find \(v(t)\) keeping in mind that the mass \(m\) is a function of \(t\) (b) Suppose that the fuel accounts for \(80 \%\) of the initial mass of the rocket and that all of the fuel is consumed in 100 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. \([\)Take \(g=9.8\) \(\left.\mathrm{m} / \mathrm{s}^{2} \text { and } c=2500 \mathrm{m} / \mathrm{s} .\right]\)

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\) \(\left(1+x^{4}\right) \frac{d y}{d x}=\frac{x^{3}}{y}\)

A tank with a 1000 -gal capacity initially contains 500 gal of water that is polluted with 50 lb of particulate matter. At time \(t=0 .\) pure water is added at a rate of \(20 \mathrm{gal} / \mathrm{min}\) and the mixed solution is drained off at a rate of 10 gal/min. How much particulate matter is in the tank when it reaches the point of overflowing?

(a) Suppose that a quantity \(y=y(t)\) increases at a rate that is proportional to the square of the amount present, and suppose that at time \(t=0 .\) the amount present is \(y_{0}\) Find an initial-value problem whose solution is \(y(t)\) (b) Suppose that a quantity \(y=y(t)\) decreases at a rate that is proportional to the square of the amount present, and suppose that at a time \(t=0 .\) the amount present is \(y_{0}\) Find an initial-value problem whose solution is \(y(t)\)

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