(a) For what values of r does the function y=e^r x satisfy...

(a) For what values of $r$ does the function $y=e^{r x}$ satisfy the differential equation $2 y^{\prime \prime}+y^{\prime}-y=0 ?$ (b) If $r_{1}$ and $r_{2}$ are the values of $r$ that you found in part (a) show that every member of the family of functions $y=a e^{r_{1} x}+b e^{r_{2} x}$ is also a solution.

(a) For what values of $r$ does the function $y=e^{r x}$ satisfy the differential equation $2 y^{\prime \prime}+y^{\prime}-y=0 ?$

(b) If $r_{1}$ and $r_{2}$ are the values of $r$ that you found in part (a) show that every member of the family of functions $y=a e^{r_{1} x}+b e^{r_{2} x}$ is also a solution.

Key Concepts

Exponential Function Solutions

Exponential functions, such as y = e^(rx), are often used as trial solutions in linear differential equations because their derivatives retain the form of the original function. This makes substituting them into the differential equation straightforward, reducing the problem to an algebraic equation in terms of the parameter r.

Characteristic Equation

When substituting an exponential trial solution into a linear homogeneous differential equation with constant coefficients, the equation simplifies to a polynomial in r known as the characteristic equation. The roots of this polynomial determine the specific exponential solutions that satisfy the differential equation.

Second-Order Linear Homogeneous Differential Equations

These equations involve a function and its second derivative with constant coefficients, and they take the general form a·y'' + b·y' + c·y = 0. Their solution methods typically involve finding the roots of the corresponding characteristic equation, which leads to a general solution expressed as a linear combination of independent solutions.

Superposition Principle

The superposition principle states that if two functions are solutions to a linear homogeneous differential equation, then any linear combination of these functions is also a solution. This principle is key to forming the general solution of the differential equation once a set of independent solutions is known.

Transcript

00:01 In this video, we're going to be looking at verifying solutions to a differential equation.

00:05 So the first thing we are x to do is find all values of r, as seen in this equation here.

00:13 This is what defines r.

00:14 Y equals e to the rx, such that y is a solution to the differential equation seen here.

00:22 So in order to do this, what we're going to do is just plug this solution into the differential equation and see what values of r will make it a true statement.

00:33 So we know y, and so we can plug y directly in, but we need to find y prime and y double prime.

00:43 So let's go ahead and do that.

00:45 So let's start with y prime.

00:49 Oops, y prime is going to equal r, e to the rx.

00:58 Then if we take the derivative of that again to find y double prime, we will get y double prime equals r squared e to the rx.

01:09 So if we take those and our expression for y and plug those in, what we will end up with is 2 times r squared e to the rx plus r e to the r x minus e to the r x.

01:36 And that all equals 0.

01:40 So let's go ahead and simplify this.

01:42 We get 2r squared, e to the rx, plus r, e to the rx minus e to the rx equals 0.

01:53 So there's an e to the rx in all three of these terms.

01:56 We can go ahead and cancel that out.

01:59 And now we're left with 2r squared plus r minus 1 equals 0.

02:10 So now in order to solve this equation for r, we can do that by factoring.

02:18 So if we go ahead and factor this out, we get r plus 1 times 2r minus 1 equals 0.

02:27 So we can now set these factors equal to 0, solve for r, and get r solutions to this equation.

02:34 So r plus 1 equals 0.

02:38 So if we solve that, we get r equals negative 1.

02:44 And let's set the other factor equal to 0 to r minus 1 equals 0.