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What kind of differential equations can be solved by direct integration?

Short Answer

Expert verified
Question: Solve the first-order differential equation dy/dx = (2x)/(y^2) using direct integration. Answer: Step 1: Identify the Differential Equation Form The given differential equation is dy/dx = (2x)/(y^2), which is already in the form dy/dx = (1/g(y)) * (f(x)) with f(x) = 2x and g(y) = y^2. Step 2: Separate Variables Separate the variables by multiplying both sides by g(y) and dx: y^2 dy = 2x dx Step 3: Integrate Both Sides Integrate the left side with respect to y and the right side with respect to x: ∫y^2 dy = ∫2x dx Step 4: Solve for the Dependent Variable (y) Find the antiderivatives for both sides of the equation: (1/3)y^3 = x^2 + C Now, solve for y in terms of x (if possible) by taking the cube root of both sides: y(x) = (3(x^2 + C))^(1/3) This is the general solution for the given differential equation.

Step by step solution

01

Identify the Differential Equation Form

A first-order differential equation that can be solved by direct integration has the following form: dy/dx = (1/g(y)) * (f(x)) where g(y) and f(x) are continuous functions of y and x respectively.
02

Separate Variables

Rearrange the equation so that the terms involving x are on the right side and terms involving y are on the left side. This can be achieved by multiplying both sides by g(y) and dx: g(y) dy = f(x) dx
03

Integrate Both Sides

Now that the variables are separated, we can integrate both sides of the equation. Integrate the left side with respect to y, and the right side with respect to x: ∫g(y) dy = ∫f(x) dx
04

Solve for the Dependent Variable (y)

Find the antiderivatives for both sides of the equation and write the general solution: G(y) = F(x) + C where G(y) and F(x) are the antiderivatives of g(y) and f(x) respectively, and C is the constant of integration. Finally, solve for y in terms of x (if possible) to obtain the solution of the differential equation.

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Key Concepts

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

Direct Integration
Direct integration is a method used to solve certain types of differential equations directly by finding their antiderivatives. Unlike more complex methods, this approach is straightforward and requires the equation to be rearranged in a specific form, typically depicting a direct relationship between the derivative of a function and its variables.

In general, a differential equation suitable for direct integration is written as:
  • \(\frac{dy}{dx} = h(x)\)
Upon identification, the solution involves integrating both sides with respect to their respective variables, effectively reversing the differentiation process.

This approach is particularly useful when the equation is simple and does not require conditions such as boundary values or initial conditions to be integrated. By using direct integration, we find the antiderivative of the function, capturing all potential solutions in a family of functions distinguished by a constant of integration \(C\).
Separation of Variables
Separation of variables is a powerful technique for solving certain types of first-order differential equations. This method involves manipulating the equation until all terms involving the dependent variable (usually y) are on one side of the equation, and all terms involving the independent variable (usually x) are on the other side.

The process looks like this:
  • Start with an equation of the form:\(\frac{dy}{dx} = \frac{1}{g(y)} \cdot f(x)\).
  • Separate variables to achieve:\(g(y) \, dy = f(x) \, dx\).
This allows us to integrate each side independently. The key benefit of this approach is the direct path it provides to the solution by ensuring that each variable is independently integrated, thus preserving the integrity of the original equation.
First-order Differential Equation
A first-order differential equation involves derivatives of a function that are of the first degree, meaning they only involve first derivatives. These equations play an essential role in modeling real-world phenomena where rates of change are dependent on a single variable at a time.

The general form of a first-order differential equation is:
  • \(\frac{dy}{dx} = f(x, y)\)
Solving such equations often involves methods like direct integration or separation of variables when applicable. In more complex scenarios, other tools must be employed. First-order differential equations are foundational in calculus due to their straightforward nature and direct applicability in many fields such as physics, biology, and economics.
Continuous Functions
Continuous functions are crucial in differential equations as they ensure smoothness and integrability. A function is considered continuous if, intuitively, you can draw it without lifting your pencil off the paper. Mathematically, a function \(f(x)\) is continuous at a point \(x = a\) if the limit of the function as it approaches \(a\) equals the function's value at \(a\).

In the context of direct integration and separation of variables, why is continuity important?
  • Continuous functions allow for the existence of derivatives and antiderivatives, which are critical for the integration process.
  • They ensure the integration follows the fundamental theorem of calculus, connecting the processes of differentiation and integration seamlessly.
  • With continuity, the solutions to differential equations are more predictable and behave consistently over their domains.
Therefore, the presence of continuous functions \(g(y)\) and \(f(x)\) is vital when employing techniques like direct integration or separation of variables.

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Most popular questions from this chapter

A 2-kW resistance heater wire whose thermal conductivity is \(k=10.4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot \mathrm{R}\) has a radius of \(r_{o}=0.06\) in and a length of \(L=15\) in, and is used for space heating. Assuming constant thermal conductivity and one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem during steady operation. Do not solve.

A long homogeneous resistance wire of radius \(r_{o}=\) \(0.6 \mathrm{~cm}\) and thermal conductivity \(k=15.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is being used to boil water at atmospheric pressure by the passage of electric current. Heat is generated in the wire uniformly as a result of resistance heating at a rate of \(16.4 \mathrm{~W} / \mathrm{cm}^{3}\). The heat generated is transferred to water at \(100^{\circ} \mathrm{C}\) by convection with an average heat transfer coefficient of \(h=3200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wire, \((b)\) obtain a relation for the variation of temperature in the wire by solving the differential equation, and \((c)\) determine the temperature at the centerline of the wire.

A spherical container of inner radius \(r_{1}=2 \mathrm{~m}\), outer radius \(r_{2}=2.1 \mathrm{~m}\), and thermal conductivity \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). The container is gaining heat by convection from the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the inner surface temperature of the container to be \(0^{\circ} \mathrm{C},(a)\) express the differential equation and the boundary conditions for steady one- dimensional heat conduction through the container, \((b)\) obtain a relation for the variation of temperature in the container by solving the differential equation, and \((c)\) evaluate the rate of heat gain to the iced water.

Consider a steam pipe of length \(L=35 \mathrm{ft}\), inner radius \(r_{1}=2\) in, outer radius \(r_{2}=2.4\) in, and thermal conductivity \(k=8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). Steam is flowing through the pipe at an average temperature of \(250^{\circ} \mathrm{F}\), and the average convection heat transfer coefficient on the inner surface is given to be \(h=\) \(15 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). If the average temperature on the outer surfaces of the pipe is \(T_{2}=160^{\circ} \mathrm{F},(a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, \((b)\) obtain a relation for the variation of temperature in the pipe by solving the differential equation, and \((c)\) evaluate the rate of heat loss from the steam through the pipe.

A solar heat flux \(\dot{q}_{s}\) is incident on a sidewalk whose thermal conductivity is \(k\), solar absorptivity is \(\alpha_{s}\), and convective heat transfer coefficient is \(h\). Taking the positive \(x\) direction to be towards the sky and disregarding radiation exchange with the surroundings surfaces, the correct boundary condition for this sidewalk surface is (a) \(-k \frac{d T}{d x}=\alpha_{s} \dot{q}_{s}\) (b) \(-k \frac{d T}{d x}=h\left(T-T_{\infty}\right)\) (c) \(-k \frac{d T}{d x}=h\left(T-T_{\infty}\right)-\alpha_{s} \dot{q}_{s}\) (d) \(h\left(T-T_{\infty}\right)=\alpha_{s} \dot{q}_{s}\) (e) None of them

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