Problem 13 Verify that $y_{1}(t)=t$ and \... [FREE SOLUTION]

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Differential Equations with Boundary Value Problems

John Polking, Al Boggess, David Arnold

$Math Studyset 91影视 Explanations$ Math

2 Edition

Chapter 4: Problem 13

Verify that $y_{1}(t)=t$ and $y_{2}(t)=t^{-3}$ are solutions to the homogeneous equation $$ t^{2} y^{\prime \prime}(t)+3 t y^{\prime}(t)-3 y(t)=0 . $$

Short Answer

Expert verified

Both $ y_1(t) = t $ and $ y_2(t) = t^{-3} $ satisfy the homogeneous equation.

Step by step solution

Calculate First Derivatives

First, we need to find the derivatives of both given functions. For $ y_1(t) = t $, the first derivative is $ y_1'(t) = 1 $.For $ y_2(t) = t^{-3} $, the first derivative is $ y_2'(t) = -3t^{-4} $.

Calculate Second Derivatives

Next, compute the second derivatives for both functions. For $ y_1(t) = t $, the second derivative is $ y_1''(t) = 0 $.For $ y_2(t) = t^{-3} $, the second derivative is $ y_2''(t) = 12t^{-5} $.

Verify Solution for $ y_1(t) = t $

Substitute $ y_1(t) = t $ and its derivatives into the differential equation:\[ t^2(0) + 3t(1) - 3(t) = 0. \]Simplified, this becomes $ 3t - 3t = 0 $, which confirms $ y_1(t) $ is a solution as the left-hand side equals zero.

Verify Solution for $ y_2(t) = t^{-3} $

Substitute $ y_2(t) = t^{-3} $ and its derivatives into the differential equation:\[ t^2(12t^{-5}) + 3t(-3t^{-4}) - 3(t^{-3}) = 12t^{-3} - 9t^{-3} - 3t^{-3}. \]Simplified, this becomes $ 0 $ as $ 12t^{-3} - 9t^{-3} - 3t^{-3} = 0 $. Thus, $ y_2(t) $ is a solution as the left-hand side equals zero.

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Key Concepts

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

Homogeneous Equation

A homogeneous equation is a type of differential equation where the function and its derivatives all relate to each other without any additional terms, meaning everything equals zero. In mathematical terms, the general form is:

The expression on the left depends on the function and its derivatives.
There are no external functions or constants on the right; it's completely set to zero.

In this context, the equation given is: \[ t^{2} y^{\prime \prime}(t) + 3t y^{\prime}(t) - 3 y(t) = 0. \] This equation is called homogeneous because every term involves the function $y(t)$ or its derivatives, and the equation equals zero. Understanding this structure can help recognize what solutions might look like鈥攐ften combinations of exponentials or power functions that balance perfectly to maintain that zero.

First Derivative

Derivatives provide the rate at which a function changes. Calculating the first derivative is crucial for analyzing how functions respond within a differential equation.For our given functions:

The first derivative of $ y_1(t) = t $ is $ y_1'(t) = 1 $. This tells us that the rate of change is constant at 1, essentially the definition of a slope in a linear function.
The first derivative of $ y_2(t) = t^{-3} $ is $ y_2'(t) = -3t^{-4} $. Here, the derivative is negative, which signifies that $ y_2(t) $ decreases as $ t $ increases.

These derivatives are necessary components in verifying if the given functions satisfy the differential equation.

Second Derivative

The second derivative gives us the rate of change of the first derivative, providing insights into the curvature or concavity of the function.With our specific functions:

For $ y_1(t) = t $, the second derivative is $ y_1''(t) = 0 $. A zero second derivative indicates no curvature; the function is a straight line.
For $ y_2(t) = t^{-3} $, the second derivative comes out to be $ y_2''(t) = 12t^{-5} $. This positive second derivative means $ y_2(t) $ is concave up, increasing more rapidly as $ t $ decreases.

Understanding these derivatives allows us to correctly substitute them back into the original equation for verification purposes.

Solution Verification

Verifying solutions to a differential equation involves checking if substituting the function and its derivatives back into the equation results in a true statement.Here's how we do it:

For $ y_1(t) = t $: We plug it into our equation: \[ t^2(0) + 3t(1) - 3(t) = 0 \]This simplifies to $ 3t - 3t = 0 $. Hence, it confirms that $ y_1(t) = t $ satisfies the equation.
For $ y_2(t) = t^{-3} $: Substituting in gives: \[ 12t^{-3} - 9t^{-3} - 3t^{-3} = 0 \]This again simplifies to zero, verifying $ y_2(t) = t^{-3} $ as a solution.

Through verification, we ensure these are indeed correct solutions, exemplifying their validity in satisfying the original differential equation.

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91影视

Verify that \(y_{1}(t)=t\) and \(y_{2}(t)=t^{-3}\) are solutions to the homogeneous equation $$ t^{2} y^{\prime \prime}(t)+3 t y^{\prime}(t)-3 y(t)=0 . $$

Short Answer

Step by step solution

Calculate First Derivatives

Calculate Second Derivatives

Verify Solution for \( y_1(t) = t \)

Verify Solution for \( y_2(t) = t^{-3} \)

Key Concepts

Homogeneous Equation

First Derivative

Second Derivative

Solution Verification

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