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Determine all values of the constant $r$ such that the given function solves the given differential equation.$$y(x)=x^{r}, \quad x^{2} y^{\prime \prime}+5 x y^{\prime}+4 y=0$$.

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Calculus 2 / BC

Chapter 1

First-Order Differential Equations

Section 2

Basic Ideas and Terminology

Differential Equations

Oregon State University

Harvey Mudd College

Baylor University

Idaho State University

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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Determine all values of th…

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Determine the values of th…

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Determine the values of $r…

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Find the general solution …

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Solve the given differenti…

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Substitute $y=e^{r x}$ int…

in this problem were given a differential equation, and we went to decide for What are this y that salts this differential equation so we can go ahead and take the derivative of this way of X, get why prime of X is equal to or just using the power will here. So r times X to the R minus one. And we get why double prime of X is equal. Teoh R minus one times R Times X to the R minus two again just using the power roll on it. So now we have expressions for by of X y prime of X y double prime of X, and we can plug them into this differential equation and see for what are we get? Zero. So let's do that. So we'll start with X squared times wide, double prime. So that's X squared times R minus one times are times X to the R minus two. Put that in parentheses. Okay, um, plus five x times Why prime, which is R times X to the R minus one plus four y plus four x to the are all right. So here we can go ahead. We want to see where this is equal to zero. Let's factor out. We can factor out in each of these equation. Well, actually, before we doctor outlets distribute in here. So this is equal to R minus one and then x to the R minus, two times X squared. Sorry, I forgot this are right here. Okay. Now XLR minus two times R squared is just extra. The are so it's nice. And here we have five x times are times extra, the R minus one while extra The R minus one times x is excellent Oregon. So we get five r times X to the are then plus four x to the are we wonder when this is equal to zero. Well, now we can factor out x to the are out of each of these terms. So we get X. The are men R minus one times are plus, five are plus four. And we wonder when this is equal to zero well, extra. The R is never going to be equal to zero for any Are so we're really just concerned with this inner part. So let's rewrite this and simplify it a bit. Um, this is equal to XDR times ar minus one times ours r squared minus are plus five r plus for we wonder what this is equal to zero. Now, I'm gonna start writing over here, continuing. We get that that's equal to X to the arm. Times are squared plus four R plus four you go to when this is equal to zero. All right, so this is just a quadratic equation. Quadratic expressions. We can factor it. We yet x to the r times, R plus two square equals zero and this is equal to zero Exactly when R is equal to negative two. So when our isn't equal to negative two, then this Why of X solves this differential equation here.

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