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

Short Answer

Expert verified
Short Answer: To solve a differential equation using direct integration, the equation must be separable, meaning it can be written in the form dy/dx = f(x) * g(y), where f(x) is a function of x only and g(y) is a function of y only. To solve the equation, we rewrite it as dy/g(y) = f(x) dx, and then we integrate both sides separately. Finally, we solve for y to find the general solution.

Step by step solution

01

Identify separable differential equations

A differential equation can be solved by direct integration if it is separable. A separable differential equation is one in which the independent variable (usually x) and dependent variable (usually y) can be separated, allowing us to integrate both sides with respect to the appropriate variable.
02

Write the general form of a separable differential equation

The general form of a separable differential equation is: dy/dx = f(x) * g(y) where dy/dx is the first derivative of y with respect to x, f(x) is a function of x only, and g(y) is a function of y only.
03

Explain the direct integration method

To solve a separable differential equation using direct integration, we can follow these steps: 1. Rewrite the equation in the form: dy/g(y) = f(x) dx 2. Integrate both sides separately: ∫1/g(y) dy = ∫f(x) dx + C 3. Solve for y to find the general solution. For example, if we have the differential equation dy/dx = x * y, we can rewrite it as dy/y = x dx. Integrating both sides, we obtain ∫1/y dy = ∫x dx + C, which yields ln|y| = 1/2x^2 + C. Solving for y, we find the general solution: y(x) = ±e^(1/2x^2 + C) or y(x) = k*e^(1/2x^2), where k = ±e^C is an arbitrary constant.

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Most popular questions from this chapter

The variation of temperature in a plane wall is determined to be $T(x)=110 x-48 x\( where \)x\( is in \)\mathrm{m}\( and \)T\( is in \){ }^{\circ} \mathrm{C}$. If the thickness of the wall is \(0.75 \mathrm{~m}\), the temperature difference between the inner and outer surfaces of the wall is (a) \(110^{\circ} \mathrm{C}\) (b) \(74^{\circ} \mathrm{C}\) (c) \(55^{\circ} \mathrm{C}\) (d) \(36^{\circ} \mathrm{C}\) (e) \(18^{\circ} \mathrm{C}\)

What is the difference between an algebraic equation and a differential equation?

What is heat generation? Give some examples.

Consider a \(1.5-\mathrm{m}\)-high and \(0.6-\mathrm{m}\)-wide plate whose thickness is \(0.15 \mathrm{~m}\). One side of the plate is maintained at a constant temperature of \(500 \mathrm{~K}\), while the other side is maintained at \(350 \mathrm{~K}\). The thermal conductivity of the plate can be assumed to vary linearly in that temperature range as \(k(T)=k_{0}(1+\beta T)\) where \(k_{0}=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and $\beta=8.7 \times 10^{-4} \mathrm{~K}^{-1}$. Disregarding the edge effects and assuming steady onedimensional heat transfer, determine the rate of heat conduction through the plate.

A spherical shell with thermal conductivity \(k\) has inner and outer radii of \(r_{1}\) and \(r_{2}\), respectively. The inner surface of the shell is subjected to a uniform heat flux of \(\dot{q}_{1}\), while the outer surface of the shell is exposed to convection heat transfer with a coefficient \(h\) and an ambient temperature \(T_{\infty}\). Determine the variation of temperature in the shell wall, and show that the outer surface temperature of the shell can be expressed as $T\left(r_{2}\right)=\left(\dot{q}_{1} / h\right)\left(r_{1} / r_{2}\right)^{2}+T_{\infty}$

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