Question

In each of Problems 19 and 20 determine the values of \(r\) for which the given differential equation has solutions of the form \(y=t^{\prime}\) for \(t>0 .\) $$ t^{2} y^{\prime \prime}+4 t y^{\prime}+2 y=0 $$

Short answer

Answer: The values of r are \(r = -1\) and \(r = -2\).

Step-by-step solution

Substitute the given function form into the differential equation

To find the values of \(r\) for which \(y = t^r\) is a solution to the given differential equation, we will plug the function \(y(t) = t^r\) into the equation and then solve for \(r\). We will also need the first and second derivatives of \(y(t)\): First derivative: \(y'(t) = r t^{r-1}\) Second derivative: \(y''(t) = r(r - 1) t^{r-2}\) Now we substitute these derivatives and the function \(y(t)\) into the given differential equation: $$ t^2 (r(r-1)t^{r-2}) + 4t (rt^{r-1}) + 2(t^r) = 0 $$

Simplify and factor the equation

Now, we simplify and factor the terms in the equation: $$ r(r-1)t^{r} + 4rt^{r} + 2t^{r} = 0 $$ Factoring out the common term \(t^r\), we have: $$ t^r(r(r-1) + 4r + 2) = 0 $$

Find the conditions on \(r\) for the equation to hold

Since \(t > 0\), we know that \(t^r\) can never be equal to zero. Thus, the term inside the parentheses must be equal to zero for the equation to hold: $$ r(r-1) + 4r + 2 = 0 $$ This is a quadratic equation in the variable \(r\), which we can solve for \(r\).

Solve the quadratic equation for \(r\)

To solve the quadratic equation, we first simplify it: $$ r^2 - r + 4r + 2 = 0 $$ This simplifies to: $$ r^2 + 3r + 2 = 0 $$ Now, we can factor the quadratic equation: $$ (r + 1)(r + 2) = 0 $$ From this, we can find the two values of \(r\) for which the equation holds: $$ r = -1 \quad \text{or} \quad r = -2 $$ So, the given differential equation has solutions of the form \(y = t^r\) for \(r = -1\) and \(r = -2\).

Key concepts

Quadratic Equations

Quadratic equations are polynomials of degree two, usually written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. Understanding quadratic equations is essential in many areas of mathematics, including solving differential equations like the one in our exercise.

In our case, we derived a quadratic equation in the variable \( r \): \( r^2 + 3r + 2 = 0 \). This equation can be solved by factoring, which involves expressing it as a product of its factors. The factors are \( (r + 1)(r + 2) = 0 \). Setting each factor equal to zero gives the roots of the equation: \( r = -1 \) and \( r = -2 \).

The solutions to our quadratic equation indicate the values of \( r \) for which the original differential equation has solutions of the form \( y = t^r \). Quadratic equations often arise in differential equations, and solving these quadratics is a vital step toward understanding the behavior of solutions.

Derivative Calculations

Derivatives measure how a function changes as its input changes. In our differential equation, we are dealing with derivatives because they are crucial to solving such equations. Derivatives are represented by symbols like \( y' \) for the first derivative and \( y'' \) for the second derivative.

For the function \( y = t^r \), the first derivative is calculated as \( y'(t) = r t^{r-1} \). This tells us how the function changes concerning the variable \( t \). The second derivative is \( y''(t) = r(r - 1) t^{r-2} \), depicting how the rate of change itself changes.

Substituting these derivatives into our differential equation helps us simplify and eventually solve for values of \( r \). Understanding how to calculate derivatives and their significance is crucial in analyzing and solving differential equations.

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial into simpler components (factors) that, when multiplied together, reproduce the original polynomial. This technique is crucial in solving quadratic equations, as we've seen in the previous sections.

In the step-by-step solution of our exercise, we encountered a polynomial in \( r \): \( r^2 + 3r + 2 = 0 \). To solve it, we factored it into \( (r + 1)(r + 2) = 0 \).

Factoring is an essential skill because it allows us to find the roots of the polynomial quickly. These roots are answers to many mathematical problems, such as differential equations. Properly understanding and practicing factoring can make tackling complex equations much more manageable.