Chapter 3: Problem 41
Use the substitution introduced in Problem 38 in Section 3.4 to solve each of the equations \(t^{2} y^{\prime \prime}-3 t y^{\prime}+4 y=0, \quad t>0\)
Short Answer
Expert verified
Question: Solve the given linear homogeneous differential equation with variable coefficients for \(t>0\): \(t^2y''(t) - 3ty'(t) + 4y(t) = 0\). Use the substitution \(y(t) = u(\ln t)\).
Answer: The solution to the given differential equation is \(y(t) = (C_1 + C_2\ln t)(t^2)\) for \(t>0\).
Step by step solution
01
Problem 38, Section 3.4 Substitution
The given substitution is: \(y(t)=u(\ln t).\) So, we need to use this substitution to rewrite the given equation in terms of \(u\) and solve the resulting equation for \(u\).
02
Find \(y'(t)\) and \(y''(t)\)
To substitute \(y(t)=u(\ln t)\) into the given equation, first we need to find \(y'(t)\) and \(y''(t)\) by differentiating \(y(t)\).
\(y'(t) = \frac{\mathrm{d}u(\ln t)}{\mathrm{d}t} = \frac{\mathrm{d}u(\ln t)}{\mathrm{d}(\ln t)} \cdot \frac{\mathrm{d}(\ln t)}{\mathrm{d}t} = \frac{1}{t} u'(\ln t)\)
\(y''(t) = \frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{1}{t}u'(\ln t) \right) = \frac{-1}{t^2} u'(\ln t)+ \frac{1}{t^2} u''(\ln t)\)
03
Substitute \(y(t)\), \(y'(t)\), and \(y''(t)\) into the given equation
Now, we'll substitute \(y(t)=u(\ln t)\), \(y'(t)=\frac{1}{t}u'(\ln t)\), and \(y''(t)=\frac{-1}{t^2}u'(\ln t)+\frac{1}{t^2}u''(\ln t)\) into the given equation:
\(t^2\left(\frac{-1}{t^2}u'(\ln t)+\frac{1}{t^2}u''(\ln t)\right)-3ty'\left(\frac{1}{t}u'(\ln t)\right)+4y=4u(\ln t)=0\)
04
Simplify the equation
Now let us simplify the equation by factoring out the common terms and cancelling out \(t\)s.
\(-u'(\ln t)+u''(\ln t)-3u'(\ln t)+4u(\ln t)=0\)
05
Rewrite simplified equation into a 2nd order linear ordinary differential equation
Rewrite the simplified equation with original variables:
\(-u'+u''-3u'+4u=0\)
06
Combine like terms and solve the 2nd order linear ODE
Combine terms and rewrite the equation as:
\(u''-4u'+4u=0\)
This is a second-order linear homogeneous differential equation with constant coefficients. We use the characteristic equation to determine the values of `r`:
\(r^2 - 4r + 4 = 0\)
The quadratic equation factors as \((r-2)^2=0\), so we have a repeated root \(r=2\). The general solution to the equation is given by:
\(u(x) = (C_1 + C_2 x)e^{2x}\)
07
Substitute back the original variables
We need to substitute back the original variables. Recall that \(y(t)=u(\ln t)\), so we have:
\(y(t)=(C_1 + C_2 \ln t)e^{2\ln t}\)
Since \(e^{\ln t^2} = t^2\), we finally have:
\(y(t)=(C_1 + C_2\ln t)(t^2)\)
The solution to the given differential equation is \(y(t)=(C_1 + C_2\ln t)(t^2)\) for \(t>0\).
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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 tool for solving differential equations, especially when they pose complexity in their current form. It works by introducing a new function and variable that simplifies the equation. In our exercise, the substitution method transforms the given ordinary differential equation (ODE) to a simpler one. We define a new function, let's say, with specifics as mentioned in the solution above, such as \(y(t)=u(\ln t)\), where \(u\) is a function of \(\ln t\).
This substitution brings two main advantages. First, it can simplify the mathematical operations needed, like derivatives in this case, and second, it sometimes transforms the ODE into a form where known methods can be easily applied to solve it. In this example, the substitution helped to convert the given equation into a more familiar second-order linear ODE with constant coefficients, which can then be solved using established techniques.
This substitution brings two main advantages. First, it can simplify the mathematical operations needed, like derivatives in this case, and second, it sometimes transforms the ODE into a form where known methods can be easily applied to solve it. In this example, the substitution helped to convert the given equation into a more familiar second-order linear ODE with constant coefficients, which can then be solved using established techniques.
Ordinary Differential Equation
An ordinary differential equation (ODE) is an equation involving functions and their derivatives. In simple terms, it represents a relationship involving rates of change. The equation given in the exercise, \(t^{2} y''-3 t y'+4 y=0\), is a second-order linear ODE because the highest derivative is the second derivative, \(y''\), and it's linear in \(y\) and its derivatives. Typical solutions to ODEs describe the behavior of physical systems, from simple motion to the dynamics of planets. Understanding how to solve ODEs is crucial because these equations are widely used to model real-life phenomena in physics, engineering, economics, and beyond. The methods of solving them range from separation of variables to more complex ones like the substitution method used in this exercise.
Homogeneous Differential Equation
A homogeneous differential equation is characterized by the absence of stand-alone terms, meaning all terms involve the function or its derivatives. When we look at the simplified form of our equation after substitution and canceling terms, we get \(u'' - 4u' + 4u = 0\). This is homogeneous because there is no term without the function \(u\) or its derivatives. In general, if an ODE has terms that depend on the independent variable (like \(t\)) or are constants outside the function or derivatives (like \(5\) or \(3t\)), it would not be homogeneous.
Solving homogeneous differential equations often involves finding the roots of the characteristic equation, as they offer a pathway to the general solution. It's a clean process where the characteristic equation governs the form of the solution, which is often a combination of exponential functions.
Solving homogeneous differential equations often involves finding the roots of the characteristic equation, as they offer a pathway to the general solution. It's a clean process where the characteristic equation governs the form of the solution, which is often a combination of exponential functions.
Characteristic Equation
The characteristic equation is a keystone for solving linear homogeneous differential equations with constant coefficients. It provides a way to determine the form of the general solution by translating the ODE into an algebraic equation. The characteristic equation corresponding to our homogeneous differential equation is \(r^2 - 4r + 4 = 0\).
The roots of this quadratic equation, which in this case are identical and equal to \(r=2\), tell us much about the solution to the ODE. For distinct real roots, the solution is a sum of exponential functions. For repeated roots, as we have here, the solution contains both an exponential function and a term that multiplies the exponential by the variable. As a consequence, the general solution for our problem is \(u(x) = (C_1 + C_2 x)e^{2x}\), which, once we back-substitute, leads to the final solution involving \(t\). This characteristic equation technique is a fundamental part of the toolkit for any student dealing with differential equations.
The roots of this quadratic equation, which in this case are identical and equal to \(r=2\), tell us much about the solution to the ODE. For distinct real roots, the solution is a sum of exponential functions. For repeated roots, as we have here, the solution contains both an exponential function and a term that multiplies the exponential by the variable. As a consequence, the general solution for our problem is \(u(x) = (C_1 + C_2 x)e^{2x}\), which, once we back-substitute, leads to the final solution involving \(t\). This characteristic equation technique is a fundamental part of the toolkit for any student dealing with differential equations.
