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Chapter 2: Problem 135

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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Most popular questions from this chapter

The variation of temperature in a plane wall is determined to be \(T(x)=52 x+25\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the temperature at one surface is \(38^{\circ} \mathrm{C}\), the thickness of the wall is (a) \(0.10 \mathrm{~m}\) (b) \(0.20 \mathrm{~m}\) (c) \(0.25 \mathrm{~m}\) (d) \(0.40 \mathrm{~m}\) (e) \(0.50 \mathrm{~m}\)

Heat is generated in a long wire of radius \(r_{o}\) at a constant rate of \(\dot{e}_{\text {gen }}\) per unit volume. The wire is covered with a plastic insulation layer. Express the heat flux boundary condition at the interface in terms of the heat generated.

Consider a large plane wall of thickness \(L=0.05 \mathrm{~m}\). The wall surface at \(x=0\) is insulated, while the surface at \(x=L\) is maintained at a temperature of \(30^{\circ} \mathrm{C}\). The thermal conductivity of the wall is \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and heat is generated in the wall at a rate of \(\dot{e}_{\text {gen }}=\dot{e}_{0} e^{-0.5 x / L} \mathrm{~W} / \mathrm{m}^{3}\) where \(\dot{e}_{0}=8 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and (c) determine the temperature of the insulated surface of the wall.

Consider a solid cylindrical rod whose ends are maintained at constant but different temperatures while the side surface is perfectly insulated. There is no heat generation. It is claimed that the temperature along the axis of the rod varies linearly during steady heat conduction. Do you agree with this claim? Why?

In a food processing facility, a spherical container of inner radius \(r_{1}=40 \mathrm{~cm}\), outer radius \(r_{2}=41 \mathrm{~cm}\), and thermal conductivity \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used to store hot water and to keep it at \(100^{\circ} \mathrm{C}\) at all times. To accomplish this, the outer surface of the container is wrapped with a 800 -W electric strip heater and then insulated. The temperature of the inner surface of the container is observed to be nearly \(120^{\circ} \mathrm{C}\) at all times. Assuming 10 percent of the heat generated in the heater is lost through the insulation, \((a)\) 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 material by solving the differential equation, and \((c)\) evaluate the outer surface temperature of the container. Also determine how much water at \(100^{\circ} \mathrm{C}\) this tank can supply steadily if the cold water enters at \(20^{\circ} \mathrm{C}\).
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