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Which of the following functions are solutions of the differential equation $ y^{"} + y = \sin x ? $

(a) $ y = \sin x $(b) $ y = \cos x $(c) $ y = \frac {1}{2} x \sin x $(d) $ y = - \frac{1}{2} x \cos x $

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$\mathrm{LHS}=y^{\prime \prime}+y=-\frac{1}{2}(-x \cos x-2 \sin x)+\left(-\frac{1}{2} x \cos x\right)=\sin x=\mathrm{RHS},$ so $y=-\frac{1}{2} x \cos x$ is a solution of the differential equation.

Calculus 2 / BC

Chapter 9

Differential Equations

Section 1

Modeling with Differential Equations

Missouri State University

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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So i want to know which of the following functions: are solutions to the differential equation. So this is our differential equation, so we want to find our y double prime in our y and plug those into it. So, let's start with a so. We need to find y double prime, so first, let's remember that the derivative of cosine is negative sine x and the derivative of sine is positive, cosine x, so there just some identity to remember. So, let's take our derivative of sin. We'Re going to get cosine x and if we take the derivative of that we get negative sine x, so we'r just gonna plug those into our equation here. So we'll have negative sine x, plus sine x, is equal to sine x m our left hand side. This is 0 and this is sine x. These are not equal. Therefore, we know a is not our answer. For b, we're gonna do the same thing: we're going to find y prime, which is negative sine x, and then we're going to get our y double pi, which would be bring down our negative derivative sine is cosine, so negative, cosine x, we plug those into our Equation, we'll have negative cosine x, plus cosine x, and bring down our right side, so our left side is 0 and my right side is is sine x and these are not the same. Therefore, is not our answer choice either force we need to find y prime, then y, double prime and plug that into our equation, so to find y prime we're going to use product rule, so we have the first time the derivative of the second plus the second Times the derivative of the first, we simplify that we'll have 1 half x, cosine x, plus 1 half sine x, i'm going to do the same thing to find a double prime. We have to use product rule for this portion here, so we take the first times the derivative of the second plus the second times. The derivative of the first is just 1 half sorry and then we still need to bring down this portion here, but it's deriand that derivative is 1 half cosine x. So if we simplify this, we'll have negative 1 half x sine x, plus 1, half cosine x, plus 1 half cosine x, dashes x, so we'll just put a 1 here or understood 1 there. So now we have our prom equation and our y equation and we can just plug those into our function, so have negative 1 half x sine x, plus cosine, x, plus r y equation and bring down our right hand side. So when we simplify, we have a negative 1 half here and a positive here, so it also just cancel with each other and we have cosine x equals sine x. However, this is not true. Therefore, c is not our answer choice either. If i can look at option d, so even though we know that this has to be the answer choice, we can still work it out. So we have our y function. We need to find our y prime, so we take derivative. We take the first times derivative of the second plus the second times the derivative of the first to simplify this we'll have positive 1 half x sine x, minus 1, half cosine x, and we use this to find our y double prime equation. So we take the first times the derivative, the second plus the second times the derivative of the first, and that was the derivative of this portion. Now we need to take the root of this portion, so we have a minus 1 half sine x, but it changes sign because the derivative cosine is negative sign so to simplify this we'll have a 1 half x, cosine x, and then this will be 1 half Plus 1 half that's just 1, so plus sine x. So now we have all 3 equations that we need to plug into our differential equation here. So i'm going to plug those in so we have 1 half x, cosine x, plus sine x, plus a negative 1 half x, cosine x and bring down our right hand side. If we look here- and here they subtract out and were left with sine x, is equal to sine x. Therefore, d is our answer: choice.

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