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Chapter 5: Problem 8

In general, differential equations of the form $$ \dot{x}=P(t)+Q(t) x+R(t) x^{2} \quad \text { (Riccati's equation) } $$ can only be solved numerically. But if we know a particular solution \(u=u(t)\), the substitution \(x=u+1 / z\) will transform the equation into a linear differential equation for \(z\) as a function of \(t\). Prove this, and apply it to \(t \dot{x}=x-(x-t)^{2}\).

Short Answer

Expert verified
The substitution transforms the given Riccati equation into the linear differential equation \( \frac{dz}{dt} + (2t)z = -\frac{1}{t} \).

Step by step solution

01

- Identify the given Riccati equation

The initial differential equation is given by \[ \frac{dx}{dt} = P(t) + Q(t)x + R(t)x^2 \].
02

- Substitute the given solution

Given the particular solution is \(u(t)\), substitute \(x = u + \frac{1}{z}\) where \(z\) is a function of \(t\).
03

- Compute the derivative of \(x\)

Differentiating \(x = u + \frac{1}{z}\) with respect to \(t\) gives \[ \frac{dx}{dt} = \frac{du}{dt} - \frac{1}{z^2} \frac{dz}{dt} \].
04

- Substitute into the Riccati equation

Substitute \(x = u + \frac{1}{z}\) and \(\frac{dx}{dt} = \frac{du}{dt} - \frac{1}{z^2} \frac{dz}{dt}\) back into the original Riccati equation gives \[ \frac{du}{dt} - \frac{1}{z^2} \frac{dz}{dt} = P(t) + Q(t) \bigg(u + \frac{1}{z}\bigg) + R(t) \bigg(u + \frac{1}{z}\bigg)^2 \].
05

- Simplify the substituted equation

Using the fact that \(u(t)\) is a particular solution, it satisfies the equation \[ \frac{du}{dt} = P(t) + Q(t)u + R(t)u^2 \].Subtract this equation from the previously substituted one: \[ -\frac{1}{z^2} \frac{dz}{dt} = Q(t) \frac{1}{z} + R(t) \bigg(2u \frac{1}{z} + \frac{1}{z^2}\bigg) \].
06

- Isolate the linear differential equation

After simplifying, we get \[ -\frac{1}{z^2} \frac{dz}{dt} = Q(t) \frac{1}{z} + 2R(t)u \frac{1}{z} + R(t) \frac{1}{z^2} \].Multiply through by \(z^2\) to get \[ -\frac{dz}{dt} = (Q(t) + 2R(t)u)z + R(t) \].Rearrange into standard linear form \[ \frac{dz}{dt} + (Q(t) + 2R(t)u)z = -R(t) \].
07

- Apply to the specific equation

Apply the transformation to the specific differential equation \( t \frac{dx}{dt} = x - (x - t)^2 \).Rewrite it in the standard Riccati form: \[ \frac{dx}{dt} = \frac{1}{t}x - \frac{1}{t}(x^2 - 2tx + t^2) = \frac{1}{t}x - \frac{1}{t}x^2 + 2x - t \].
08

- Identify a particular solution

Checking for a simple solution, one can see that \(u(t) = t\) works since \[ t \frac{d}{dt}(t) = t - (t - t)^2 = t \].
09

- Substitute the solution and derive the linear equation

Using \(x = t + \frac{1}{z}\) and follow steps 4-6 to transform and simplify results in: \[ -\frac{dz}{dt} = (\frac{1}{t} + 2*t - \frac{1}{t})z + \frac{1}{t} \].
10

- Final simplified linear equation

The final linear differential equation is \[ \frac{dz}{dt} + (2t)z = -\frac{1}{t} \].

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Key Concepts

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

particular solution
A particular solution to a differential equation is a specific solution that satisfies the equation. In the context of a Riccati differential equation, a particular solution, denoted as \(u = u(t)\), is used to transform the nonlinear equation into a linear one. To find this particular solution, we often test simple functions. For the given problem, trying \(u(t) = t\) works, as it satisfies the equation when substituted back.
substitution method
In solving a Riccati differential equation, the substitution method is a powerful tool. Here, we use a particular solution, \(u = u(t)\), to make the substitution \(x = u + \frac{1}{z}\). This transforms the original nonlinear differential equation into a linear one. This method is incredibly useful because linear equations are generally easier to solve.
linear differential equation
A linear differential equation has the general form \(a(t) \frac{dz}{dt} + b(t)z = c(t)\). This form is much simpler to solve compared to nonlinear ones. For example, after substituting \(x = u + \frac{1}{z}\) into the original Riccati equation \(\frac{dx}{dt} = P(t) + Q(t)x + R(t)x^2\) and simplifying, we derived the linear equation \(\frac{dz}{dt} + (Q(t) + 2R(t)u)z = -R(t)\). This reveals the structure we can work with for further solutions.
transformation of equations
The process of transforming equations is essential in solving complex differential equations. For Riccati equations, knowing a particular solution allows us to transform the nonlinear equation into a linear form. This transformation is achieved through a careful substitution of variables. For example, substituting \(x = u + \frac{1}{z}\) where \(u = u(t)\) is the particular solution, modified the original nonlinear structure into a linear differential equation, making it easier to analyze and solve. The transformed equation is \[ \frac{dz}{dt} + (Q(t) + 2R(t)u)z = -R(t) \]

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Most popular questions from this chapter

For the differential equation \(\dot{x}=2 t x+t\left(1+t^{2}\right)\), show that the solution \(x(t)\) that passes through \((t, x)=(0,0)\) has a local minimum at \(t=0\).

Solve the differential equation \((1+t x) \dot{x}=-x^{2}\). (Hint: Try the substitution \(w=t x\).)

Show that \(x=C t^{2}\) satisfies the differential equation \(t \dot{x}=2 x\) for all values of the constant \(C\). But all the corresponding solution curves pass through the point \((0,0)\). How do you reconcile this observation with Theorem 5.8.1?

(a) Consider the equation $$ t \dot{x}=x-f(t) x^{2}, \quad t>0 $$ where \(f\) is a given continuous function, and \(x=x(t)\) is the unknown function. Show that the substitution \(x=t z\) transforms it into a separable equation in \(z=z(t)\). (b) Let \(f(t)=t^{3} /\left(t^{4}+2\right)\) and find the solution curve through the point \((1,1)\).

Solve the differential equation \(1+(2+t / x) \dot{x}=0, t>0, x>0\)
See all solutions

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