Problem 8 In Problems 1 through 16, a homo... [FREE SOLUTION]

Chapter 3: Problem 8

In Problems 1 through 16, a homogeneous second-order linear differential equation, two functions $y_{1}$ and $y_{2}$, and a pair of initial conditions are given. First verify that $y_{1}$ and $y_{2}$ are solutions of the differential equation. Then find a particular solution of the form $y=c_{1} y_{1}+c_{2} y_{2}$ that satisfies the given initial conditions. Primes denote derivatives with respect to $x$. $$ y^{\prime \prime}-3 y^{\prime}=0 ; y_{1}=1, y_{2}=e^{3 x} ; y(0)=4, y^{\prime}(0)=-2 $$

Short Answer

Expert verified

The particular solution is $y = \frac{14}{3} - \frac{2}{3}e^{3x}$.

Step by step solution

Verify Solutions

Substitute $y_1 = 1$ into the differential equation $y'' - 3y' = 0$. For $y_1 = 1$, the derivatives are $y_1' = 0$ and $y_1'' = 0$. Hence, substituting gives $0 - 3*0 = 0$, which verifies $y_1$ is a solution.Next, substitute $y_2 = e^{3x}$ with its derivatives $y_2' = 3e^{3x}$ and $y_2'' = 9e^{3x}$. Substitute these into the differential equation: $9e^{3x} - 3(3e^{3x}) = 0$, simplifying to $9e^{3x} - 9e^{3x} = 0$, confirming $y_2$ is a solution.

Form General Solution

The general solution for the differential equation can be written as a linear combination of the two solutions, $y = c_1y_1 + c_2y_2$. Substituting $y_1 = 1$ and $y_2 = e^{3x}$, the general solution becomes $y = c_1 + c_2e^{3x}$.

Apply Initial Conditions

We need to find constants $c_1$ and $c_2$ that satisfy the initial conditions: $y(0) = 4$ and $y'(0) = -2$. From $y = c_1 + c_2e^{3x}$, when $x = 0$, $y(0) = c_1 + c_2 = 4$.The derivative is $y' = 0 + 3c_2e^{3x}$, so $y'(0) = 3c_2 = -2$.

Solve for Constants

From $y'(0) = 3c_2 = -2$, solve for $c_2$:\[c_2 = -\frac{2}{3}\].Substitute $c_2$ into the equation $c_1 + c_2 = 4$:\[c_1 - \frac{2}{3} = 4\]\[c_1 = 4 + \frac{2}{3} = \frac{12}{3} + \frac{2}{3} = \frac{14}{3}\].

Find Particular Solution

Substitute $c_1 = \frac{14}{3}$ and $c_2 = -\frac{2}{3}$ back into the general solution:\[y = \frac{14}{3} - \frac{2}{3}e^{3x}\].This particular solution satisfies the given initial conditions.

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Key Concepts

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

Initial Conditions

Initial conditions are a vital part of solving differential equations. They provide specific values that a solution must satisfy, effectively narrowing down the infinite possibilities of solutions to a more precise one. In the given exercise, we are provided with the initial conditions: $ y(0) = 4 $ and $ y'(0) = -2 $. These conditions specify the values of the function and its first derivative at $ x = 0 $.
Consider them as starting points or anchor points. Without them, we could not determine the constants in our solution. With any second-order differential equation, two initial conditions are commonly used because the solution usually involves two arbitrary constants.

They guide us in determining unknown constants.
They ensure the solution aligns with specific physical or theoretical situations.

Second-order Linear Differential Equation

Understanding second-order linear differential equations is crucial because they often arise in both theoretical and applied contexts. A second-order differential equation involves the second derivative of a function. In this problem, the equation given is: \[ y'' - 3y' = 0 \]This form is recognized as 'linear' because each term involving $ y $ or its derivatives is to the first power and there are no products involving $ y $ and its derivatives. Linear differential equations are often simpler to solve and analyze because they have superposition characteristics, allowing any linear combination of solutions to also be a solution.
Key characteristics include:

The highest derivative is of the second order, denoted by $ y'' $.
The terms are linear in $ y $ and its derivatives.
They can generally be approached by finding solutions that form a basis for all possible solutions, and then solving for particular solutions meeting given conditions.

Homogeneous Equation

In the context of differential equations, a homogeneous equation is one in which every term is a function of the dependent variable or its derivatives. For the given exercise, the equation $ y'' - 3y' = 0 $ is homogeneous. This means that if zero is substituted for the variable, the equation will still hold true.
The property of homogeneity implies that if $ y_1(x) $ and $ y_2(x) $ are solutions, then $ c_1y_1(x) + c_2y_2(x) $ is also a solution for any constants $ c_1 $ and $ c_2 $. This is the superposition principle. It's particularly useful because it allows constructing more complex solutions from simpler ones.

Each term in a homogeneous equation involves only the derivative(s) of the function.
The right side of the equation is zero.
Homogeneous equations often allow the construction of general solutions using linear combinations of solutions.

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Short Answer

Step by step solution

Verify Solutions

Form General Solution

Apply Initial Conditions

Solve for Constants

Find Particular Solution

Key Concepts

Initial Conditions

Second-order Linear Differential Equation

Homogeneous Equation

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