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Chapter 9: Problem 14

Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume \(C, C_{1}, C_{2}\) and \(C_{3}\) are arbitrary constants. $$u(t)=C_{1} t^{2}+C_{2} t^{3} ; t^{2} u^{\prime \prime}(t)-4 t u^{\prime}(t)+6 u(t)=0$$

Short Answer

Expert verified
Question: Verify that the function \(u(t)=C_{1} t^{2}+C_{2} t^{3}\) is a solution to the second-order differential equation \(t^{2} u^{\prime\prime}(t)-4 t u^{\prime}(t)+6 u(t)=0\). Answer: By finding the first and second derivatives of the function \(u(t)\) and substituting them into the given differential equation, we simplified the equation to \(0=0\), which confirms that the given function is indeed a solution to the differential equation.

Step by step solution

01

Find the first and second derivatives of the given function

We start by finding the first and second derivatives of the function \(u(t)=C_{1} t^{2}+C_{2} t^{3}\). Using the power rule, we get: $$ u^{\prime}(t)=\frac{d}{dt}(C_{1} t^{2}+C_{2} t^{3})=2C_{1} t+3C_{2} t^{2} $$ and $$ u^{\prime\prime}(t)=\frac{d^{2}}{dt^{2}}(C_{1} t^{2}+C_{2} t^{3})=2C_{1} +6C_{2} t. $$
02

Substitute the function and its derivatives into the differential equation

Now that we have the first and second derivatives, we will substitute them into the given differential equation: $$ t^{2} u^{\prime\prime}(t)-4 t u^{\prime}(t)+6 u(t)=0 $$ By substituting, we get: $$ t^{2}(2C_{1} +6C_{2} t)-4 t(2C_{1} t+3C_{2} t^{2})+6(C_{1} t^{2}+C_{2} t^{3})=0 $$
03

Simplify the equation

Now, we will simplify this equation by expanding, combining, and grouping the terms: $$ 2C_{1}t^{2}+6C_{2}t^{3}-8C_{1}t^{2}-12C_{2}t^{3}+6C_{1}t^{2}+6C_{2}t^{3}=0. $$ Combining the terms, we get: $$ (2C_{1}-8C_{1}+6C_{1})t^{2}+(6C_{2}-12C_{2}+6C_{2})t^{3}=0, $$ which simplifies to: $$ 0\cdot t^{2}+0\cdot t^{3}=0 $$
04

Conclusion

Since the equation simplifies to \(0=0\) after substituting the function \(u(t)\) and its derivatives, we have verified that the given function \(u(t)=C_{1} t^{2}+C_{2} t^{3}\) is indeed a solution to the differential equation: \(t^{2} u^{\prime\prime}(t)-4 t u^{\prime}(t)+6 u(t)=0\).

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Key Concepts

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

Second Derivative
The second derivative of a function is the derivative of the derivative. It provides information on the curvature of the function's graph, indicating how the rate of change of the function's slope is itself changing. In the context of the given problem, the second derivative, denoted as \(u''(t)\), was calculated from the function \(u(t)=C_{1} t^{2}+C_{2} t^{3}\). Using the power rule, which states that the derivative of \(t^n\) is \(nt^{n-1}\), we found the second derivative to be \(2C_{1} +6C_{2} t\). It's important to understand that the presence of constants \(C_{1}\) and \(C_{2}\) does not change the application of the power rule.

By finding the second derivative, we can study the acceleration of the function or, in physical contexts, motion under constant forces. In differential equations, the second derivative often represents the intrinsic properties of the system, such as natural frequencies in oscillatory systems.
Solving Differential Equations
Solving differential equations involves finding a function or a set of functions that satisfy the equation. There are various methods for solving different types of differential equations. In this case, we are dealing with a linear homogeneous second-order differential equation with constant coefficients, which often arise from problems in physics and engineering. The general approach starts with finding the function's derivatives and substituting them back into the equation.

The solution process requires understanding both the form of the differential equation and the conditions under which it is applied, such as initial values or boundary conditions. The ability to find and verify solutions like the one shown is fundamental to applying mathematical models to real-world phenomena.
Power Rule for Derivatives
The power rule is a basic rule in calculus used to find the derivative of a function that is a power of a variable. It states that if you have a function of the form \(f(x) = ax^n\), where \(a\) is a coefficient and \(n\) is a rational exponent, the derivative \(f'(x)\) is found by multiplying the exponent by the coefficient and decreasing the exponent by one. Precisely, \(f'(x) = n \times ax^{n-1}\).

This rule was utilized in the step-by-step solution to determine the first and second derivatives of \(u(t)\). It is an essential tool for students to master as it is widely used in different fields of science and engineering to solve problems involving rates of change.
Verifying Solutions
Verifying solutions to differential equations is a crucial part of the solution process. It ensures that the functions proposed do satisfy the given equation. This is done by substituting the function, and its derivatives, back into the equation and confirming that the result holds true for all variable values. In this problem, the substitution led to an identity, \(0=0\), proving that the function \(u(t)\) is indeed a solution to the differential equation.

Verification can be thought of as the 'proof' of your solution. It is a way to check your work and to ensure the integrity of the results. This step should not be skipped, as it is possible to make mistakes in the calculation of derivatives or the algebra involved in the simplification process.

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Most popular questions from this chapter

Solving initial value problems Solve the following initial value problems. $$y^{\prime}(t)=1+e^{t}, y(0)=4$$

Motion in a gravitational field An object is fired vertically upward with an initial velocity \(v(0)=v_{0}\) from an initial position \(s(0)=s_{0}\) . a. For the following values of \(v_{0}\) and \(s_{0},\) find the position and velocity functions for all times at which the object is above the ground \((s=0).\) b. Find the time at which the highest point of the trajectory is reached and the height of the object at that time. $$v_{0}=49 \mathrm{m} / \mathrm{s}, s_{0}=60 \mathrm{m}$$

The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for \(t \geq 0,\) graph the solution, and determine the first month in which the loan balance is zero. $$B^{\prime}(t)=0.0075 B-1500, B(0)=100,000$$

Chemical rate equations Let \(y(t)\) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation \(\frac{d y}{d t}=-k y^{n},\) for \(t \geq 0,\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1),\) the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\). c. Graph the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\).

A fish hatchery has 500 fish at \(t=0,\) when harvesting begins at a rate of \(b>0\) fish/year. The fish population is modeled by the initial value problem \(y^{\prime}(t)=0.01 y-b\) \(y(0)=500,\) where \(t\) is measured in years. a. Find the fish population, for \(t \geq 0,\) in terms of the harvesting rate \(b\) b. Graph the solution in the case that \(b=40\) fish/year. Describe the solution. c. Graph the solution in the case that \(b=60\) fish/year. Describe the solution.
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