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Solve the homogeneous differential equation. \(y^{\prime}=\frac{x y}{x^{2}-y^{2}}\)

Short Answer

Expert verified
The solution to the given homogeneous differential equation is \(y = \frac{K x^2 }{1 + K x}\), where K is a constant.

Step by step solution

01

Substitute y with vx

Make a substitution \(y = vx\) where \(v = \frac{y}{x}\), and replace \(y'\) with \(\frac{dv}{dx} \cdot x + v\), to turn the differential equation into a function of one variable. The given equation turns into: \(\frac{dv}{dx} \cdot x + v = \frac{x \cdot x \cdot v}{x^{2}- x^2 \cdot v^{2}}\)
02

Simplify the equation

Now, simplify the equation. It becomes: \(\frac{dv}{dx} \cdot x + v = \frac{v}{1 - v^{2}}\). Rearrange the terms to isolate \(\frac{dv}{dx}\) alone on one side, obtaining \(\frac{dv}{dx} = \frac{v-v^{2}}{x}\)
03

Separate the variables

The equation is now separable, so divide both sides by \(v-v^{2}\) and multiply by \(dx/x\) to group all x terms on one side and all v terms on the other. We get: \(\frac{dv}{v-v^{2}} = \frac{dx}{x}\)
04

Integrate both sides

Next, integrate both sides of the equation: \(\int \frac{dv}{v-v^{2}} = \int \frac{dx}{x}\). Solving these integrals yield: \( -\ln |v - 1| + \ln |v| = \ln|x| + C\), where C is the constant of integration.
05

Combine terms and solve for y

Combine the terms and exponentiate to solve for v. \( \frac{|v|}{|v - 1|} = |x| e^C \). This gives \(v = \frac{x e^C }{1 + x e^C}\). Since we substituted \(y = vx\) at the beginning, multiply v by x to get y. Thus, the solution is \(y = \frac{x^2 e^C }{1 + x e^C}\). Because \(e^C\) is also a constant, it can be replaced by another constant K: \(y = \frac{K x^2 }{1 + K x}\).

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Key Concepts

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

Substitution Method
The substitution method is a powerful technique used to solve differential equations that might not be immediately tractable. In the context of homogeneous differential equations, it involves replacing a function with another function, typically to reduce the number of variables or to convert the equation into a recognizable form. In our exercise, we replace the variable 'y' by 'vx' where 'v' is a new function expressed as \(v = \frac{y}{x}\). By doing this, we change the problem at hand into one involving only a single variable which simplifies the process.

Crucially, the choice of substitution should simplify the equation. As it turns out, by substituting \(y\) with \(vx\), and replacing \(y'\) with \(\frac{dv}{dx} \cdot x + v\), we can re-express the given differential equation in terms of 'v' and 'x', which then allows for easier manipulation and steps towards a solution. This method is commonly used since it often makes the next step, separation of variables, much simpler or even possible.
Separable Differential Equations
Separable differential equations are those that can be manipulated such that all instances of one variable and its differentials are on one side of the equation and all instances of the other variable and its differentials are on the other side. Mathematically, they take the form \(\frac{dy}{dx} = g(x)h(y)\) where 'g' is a function of 'x' only and 'h' is a function of 'y' only.

In our exercise, after applying the substitution method, we arrive at an equation that allows us to isolate 'dv' and 'dx' on opposite sides. By dividing both sides by \(v-v^{2}\) and multiplying by \(\frac{dx}{x}\), we successfully separate the variables: \(\frac{dv}{v-v^{2}} = \frac{dx}{x}\). This is a clear indication that the equation is now separable and we can proceed with integration of both sides independently. The beauty of separable equations lies in their simplicity — once separated, the path to solution involves straightforward integration.
Integrating Factors
Integrating factors are a technique used to solve linear differential equations that are not already separable. The method involves multiplying the entire equation by an aptly chosen function, known as an integrating factor, which turns the non-separable equation into one that can be integrated easily. For an equation of the form \(y' + p(x)y = q(x)\), the integrating factor is typically \(\exp(\int p(x)dx)\).

However, in the example at hand, we did not require an integrating factor because the equation became separable after the substitution method was applied, allowing us to integrate directly. Integrating factors are particularly useful when dealing with non-separable, linear first-order differential equations, and can be instrumental in simplifying complex differential equations to a form that is more easily solvable.

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Most popular questions from this chapter

Consider a tank that at time \(t=0\) contains \(v_{0}\) gallons of a solution of which, by weight, \(q_{0}\) pounds is soluble concentrate. Another solution containing \(q_{1}\) pounds of the concentrate per gallon is running into the tank at the rate of \(r_{1}\) gallons per minute. The solution in the tank is kept well stirred and is withdrawn at the rate of \(r_{2}\) gallons per minute. A 200 -gallon tank is half full of distilled water. At time \(t=0\), a solution containing \(0.5\) pound of concentrate per gallon enters the tank at the rate of 5 gallons per minute, and the well-stirred mixture is withdrawn at the rate of 3 gallons per minute. (a) At what time will the tank be full? (b) At the time the tank is full, how many pounds of concentrate will it contain?

Match the differential equation with its solution. $$ \begin{array}{ll} \underline{\text { Differential Equation }} & \underline{\text { Solution}} \\\ y^{\prime}-2 x=0 &\quad (a) y=C e^{x^{2}} (b) y=-\frac{1}{2}+C e^{x^{2}} (c) y=x^{2}+C (d) y=C e^{2 x} \end{array} $$

True or False? Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. Show that if \(y=\frac{1}{1+b e^{-k t}}\), then \(\frac{d y}{d t}=k y(1-y)\).

Solve the first-order differential equation by any appropriate method. y d x+(3 x+4 y) d y=0

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. In exponential growth, the rate of growth is constant.

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