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Chapter 11: Problem 6

Given that the differential equation $$ t^{3} \frac{d^{3} x}{d t^{3}}-(t+3) t^{2} \frac{d^{2} x}{d t^{2}}+2 t(t+3) \frac{d x}{d t}-2(t+3) x=0 $$ has two linearly independent solutions of the form \(t^{\prime}\), where \(n\) is an integer, find the general solution.

Short Answer

Expert verified
The general solution to the given differential equation can be written as: \[ x(t) = C_1 t + C_2 y(t) + C_3 z(t) \] where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants, and \(y(t)\) and \(z(t)\) are linearly independent solutions not in the form \(t^n\). We found only one linearly independent solution in the form \(t^n\), which was \(t\), but could not determine the explicit forms of \(y(t)\) and \(z(t)\).

Step by step solution

01

Identify the equation's type and any existing patterns

The given equation is a third-order, linear, homogeneous differential equation with variable coefficients.
02

Plug in the proposed solution form \(x = t^n\) into the equation and simplify it

We are given that the solutions have the form \(x = t^n\). Let's differentiate x with respect to t and plug it into the equation as follows: First derivative: \[\frac{dx}{dt} = n t^{n-1}\] Second derivative: \[\frac{d^2x}{d t^2} = n(n-1) t^{n-2}\] Third derivative: \[\frac{d^3x}{d t^3} = n(n-1)(n-2) t^{n-3}\] Now, plug these derivatives back into the original equation: \[ t^3 n(n-1)(n-2) t^{n-3} - (t+3) t^2 n(n-1) t^{n-2} + 2t(t+3) n t^{n-1} - 2(t+3) t^n = 0 \] Next, let's simplify the equation: \[ t^n n(n-1)(n-2) - t^n (t+3) n(n-1) + 2t^n n(t+3) - 2(t+3)t^n = 0 \] Factor out the common term \(t^n\): \[ t^n [n(n-1)(n-2) - (t+3) n(n-1) + 2n(t+3) - 2(t+3)] = 0 \] Since \(t^n \neq 0\), we can divide both sides by \(t^n\) and obtain the equation: \[ n(n-1)(n-2) - (t+3) n(n-1) + 2n(t+3) - 2(t+3) = 0 \]
03

Solve for the possible values of n

To solve for n, we will first set the coefficient of \(t^n\) to zero: \[ n(n-1)(n-2) - (t+3) n(n-1) + 2n(t+3) - 2(t+3) = 0 \] This is a quadratic equation in n. Solve for n by factoring: \[ (n-1)[-n^2 + (t+5)n - 2(t+3)] = 0 \] Thus, we obtain two possible values for n: 1) \(n = 1\): The first linearly independent solution. 2) Solve for n from the quadratic equation: \[-n^2 + (t+5)n - 2(t+3) = 0\] Since the second condition is not helpful in explicitly solving for n, we only have one solution in the form \(t^n\), which is \(n = 1\).
04

Write the general solution using the linearly independent solutions

Since we only found one linearly independent solution in the form \(t^n\) (i.e., \(n = 1\)), the general solution to the given differential equation can be written as: \[ x(t) = C_1 t^1 + C_2 y(t) + C_3 z(t) \] where \(C_1\), \(C_2\), and \(C_3\) are arbitrary constants, and \(y(t)\) and \(z(t)\) are linearly independent solutions not in the form \(t^n\). Thus, we have found the general solution, but we could not determine the explicit forms of \(y(t)\) and \(z(t)\).

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Key Concepts

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

Third-order differential equation
A third-order differential equation involves derivatives up to the third order of a function with respect to a variable. In our example, the function is \(x(t)\), and we differentiate it with respect to \(t\). The general form of such an equation is:
  • \(a(t)\frac{d^3 x}{dt^3} + b(t)\frac{d^2 x}{dt^2} + c(t)\frac{dx}{dt} + d(t)x = 0\)
In this equation, \(a(t), b(t), c(t), \text{and } d(t)\) are functions of \(t\).
Third-order differential equations are more complex due to the higher degree of derivatives compared to first or second order. They often model more intricate systems, such as those involving exchange of momentum or energy, which require detailed analysis and solutions.
Linear homogeneous differential equation
Linear homogeneous differential equations have the standard property that they equate to zero. Our sample equation is homogeneous since there are no terms independent of \(x(t)\) or its derivatives. The equation takes a linear form:
  • The function and its derivatives appear in the first degree.
  • There are no products of the function, its derivatives, or any non-zero constant terms on one side of the equation.
These equations are significant in mathematics and engineering because they often model systems that revert to a stable state or evolve based on their initial conditions. Getting familiar with them helps in deciphering various physical phenomena.
Variable coefficients
Variable coefficients mean the coefficients in front of the function and its derivatives depend on the variable \(t\). Our given equation demonstrates variable coefficients, for example, with the term \((t+3)t^2\). Variable coefficients add complexity as they affect how solutions are derived and analyzed.
In contrast, constant coefficients simplify the process as they often lead to solutions using straightforward techniques like characteristic equations. However, with variable coefficients, solutions can become possible by specialized methods like the method of undetermined coefficients or variation of parameters. Understanding these differences is crucial when tackling differential equations with various coefficient types.
Linearly independent solutions
Solutions to differential equations are linearly independent if one solution cannot be expressed as a linear combination of others. For our third-order equation, we aim to find three linearly independent solutions since it's of third order.
In mathematical terms, the Wronskian determinant of these solutions is not zero. This ensures that each solution contributes uniquely to the general solution of the differential equation. Therefore, the general solution is expressed as:
  • \(x(t) = C_1 y_1(t) + C_2 y_2(t) + C_3 y_3(t)\)
where \(C_1, C_2, \text{and } C_3\) are constants, and \(y_1(t), y_2(t),\) and \(y_3(t)\) are independent solutions, each integral to constructing the full set of possible solutions for the equation.

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Most popular questions from this chapter

Show that Theorem \(11.30\) does not in general apply to nonlinear equations by showing that although \(f\) such that \(f(t)=t\) is solution of $$ \frac{d^{2} x}{d t^{2}}-t^{2} \frac{d x}{d t}+x^{2}=0 $$ and \(g\) such that \(g(t)=1\) is a solution of $$ \frac{d^{2} x}{d t^{2}}-t^{2} \frac{d x}{d t}+x^{2}=1 $$ the sum \(f+g\) is not a solution of $$ \frac{d^{2} x}{d t^{2}}-t^{2} \frac{d x}{d t}+x^{2}=1 $$

Show that each of the following equations is self-adjoint and write each in the form \((11.113) .\) (a) \(t^{3} \frac{d^{2} x}{d t^{2}}+3 t^{2} \frac{d x}{d t}+x=0\). (b) \(\sin t \frac{d^{2} x}{d t^{2}}+\cos t \frac{d x}{d t}+2 x=0\). (c) \(\left(\frac{t+1}{t}\right) \frac{d^{2} x}{d t^{2}}-\frac{1}{t^{2}} \frac{d x}{d t}+\frac{1}{t^{3}} x=0\)

(a) A first-order differential equation of the form $$ \frac{d x}{d t}+a(t) x+b(t) x^{2}+c(t)=0 $$ is called a Riccati equation. Show that the transformation $$ u=\frac{P(t) \frac{d x}{d t}}{x} $$ transforms the self-adjoint second-order equation $$ \frac{d}{d t}\left[P(t) \frac{d x}{d t}\right]+Q(t) x=0 $$ into the special Riccati equation $$ \frac{d u}{d t}+\frac{1}{P(t)} u^{2}+Q(t)=0 $$ (b) Use the result of part (a) to transform the self-adjoint equation $$ \frac{d}{d t}\left[t \frac{d x}{d t}\right]+(1-t) x=0 $$ into a Riccati equation. For this Riccati equation find a solution of the form \(c t^{\prime \prime}\). Then employ the transformation of part (a) to find a solution of Equation (A). Finally, use the method of reduction of order to obtain the general solution of Equation (A).

Find the unique solution \(\phi\) of the nonhomogeneous linear system $$ \frac{d x}{d t}=\left(\begin{array}{rr} -1 & 1 \\ -12 & 6 \end{array}\right) x+\left(\begin{array}{l} 3 e^{4} \\ 8 e^{4} \end{array}\right) $$ that satisfies the initial condition \(\phi(0)=\left(\begin{array}{l}2 \\\ 4\end{array}\right)\).

Use the Sturm separation theorem to show that hetween any two consecutive zeros of \(\sin 2 t+\cos 2 t\) there is precisely one zero of \(\sin 2 t-\cos 2 t\).
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