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(a) For what values of $ r $ does the function $ y = e^{rx} $ satisfy the differential equation $ 2y^{"} + y^{'} - 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 = ae^{r_1{x}} + be^{r_2{x}} $ is also a solution.

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(a) $$\Rightarrow r=\frac{1}{2} \text { or }-1$$(b) $$=0=\mathrm{RHS}$$

Calculus 2 / BC

Chapter 9

Differential Equations

Section 1

Modeling with Differential Equations

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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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(a) For what values of $r$…

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For what values of $ r $ d…

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(a) For what values of r d…

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For what values of $r$ doe…

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(a) Show that every member…

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So you want to find values of r that satisfy the differential equation. 2 time plus y prime minus y equals 0 point. So the first step you always want to figure out what values can you plug in to this equation? So we know we're going to need a y double piminy pim. We already have our y equations, so let's find our y pin equation. So derivative of this is just r times e to the r x sinching rule. Take the derivative of what leads inside a thern exponent, which is our and multiply it by the unitin. So now we need to find y double prime. So it's important to remember that r is a constant because we're still using chain rule, not product rule, so we's going to use chain rule here. So this is just the derivative of this function, which is r some i'm going to multiply r times r and then bring down the e to the r x tati simplifies into r square e to the r x. That'S why it's important to remember that that was chain rule. The r is a constant, because otherwise you would try to use product rules, but that's not the right way. So now we have all 3 of our functions and we're gonna plug them into our original inpatient. So we have to plug in our y double prime, which is r squared to the r x plus y prime, which is r times e to the r x and minus our original equation, which is e to the r x. And we want to prove that this equals 0 so then distribute r 2 to become 2 r square e to the r x plus r e to the r x minus e to the r x equals 0. Now we see a common factor of e to the r x and all 3 of these, so we're going to factor that out and when we do that were left with 2 r, squared plus r minus 1 equals 0 point. So i only factor we need e to the r x equals 0 and we also need 2 times r, squared plus r minus 1 equals 0. Now our exponent of functions never equals 0. So we can stretch that we didn't eliminate that so now we're just looking for when 2 r, squared plus r minus 1 equals 0 point and in the nation were comfortable, you can replace the r s with x's, but essentially is the same. I want to factor by grouping, so in order to do that, we multiply the these 2. We get negative 2 and wetted to positive 1. So are factors of 2 are 1 and 2 and you have to make 1 of these negative to add up to positive 1. So are 1 is going to be negative, so with back of our grouping, we just split apart our middle term, so we have 2 x, squared plus 2 x minus 1 and then bring down our minus 1 point. So we're going to group these 2 terms and then were going to group these 2 terms and we factored out the greatest common factors. So the greatest nomonfactor of these is 2 x, x, plus 1 and thy're gonna bring out a minus 1. Here we was put x plus 1 on the inside or 2 factors are 2 x, minus 1 and x, plus 1 equals 0 point. So again, here we're going to have we're going to set both or factors equal to 0, so we have 2 x, minus 1 equals 0 and when x, plus 1 equals 0 point. So here we're gonna, add 1 on both sides. We have 2 x, equals 1 divided by 2, x, equals 1 half, and here we just attack 1 from both sides. So x equals negative 1 point, so we found our 2 r variables values of r. So when i write the 1 here are equals 1 half and negative 1 point. Those are 2 r values that satisfies the equation that we solve for here. So part b asks for us to use those values and show that every member of the family of functions of this new equation here also satisfies this equation. Here. So again, we're going a we're going to need a y double prime of y prime and our original y equation. So we know we're going to find our y prime equation y double in equation, but the first step is to present our variables are our variables. So we know that y equals a times e to the 1 half x, plus in r 1 plus b times e to the negative x im just going to put the negative leave of the 1. So we have our original equation now we need our prime equation. So we're going to take the derivative this, so this is a constant, so we just take the derivative of our exponent, which is 1 half times a times e to the 1 half and same thing here. These are constants and we just take the derivative of our upper function, which is negative. 1 said it would just be a minus b to the negative x and, lastly, we need our y double ton equation. So again, we're gonna take the derivative of this up here because you both are constants so 1 half times 1 half a e to the 1 half and then we're going to take our derivative of this inner function, because that's constant still so that would become plus B to the negative x negative times is positive so to simplify this, we have 1 fourth e to the 1 half x, plus b times e to the negative sixpoint, and now we have all 3 of our function as impudent our equation here, 1 of them to Step out, the way are we're going to plug everything in so we have 2 times our y double prime equation: we're gonna plug in 1. Fourth, a times e to the 1 half x, plus b, times e to the negative x plus our y prime equation, which is 1 half a e to the 1 half x minus b times e to the they want to plug in our y equation. And don't forget to say that equals to 0 gonna distribute so we have 1 half a times e to the 1 half x, plus 2 b, plus b b. And then you just bring everything else down because of something else to distribute and we're going to distribute his negative here. So minus a times e to the 1 half x, minus b e to the negative x equals 0. So now we have a bunch of variables and numbers so we're just going to simplify the pond the the basis. So i know that we can combine these 2, so i have a stone so we'll have a times e to the 1 half x, because 1 half plus 1 half is 1. Next. We can just break down this here so plus 2 b e to negative x. Then we can bring this down here so minus e to negative x because the bind us i want to combine these. So that's 2. Minus 1 is just give us a positive 1 here, so we have a plus b times e to the negative x and then lastly, you're gonna bring this down. So we have a minus a times e to the 1 half and then we can bring that down to that, be to the negative x equals 0 and as you see, we have a positive and negative those cancel and a positive and negative or those cancel. So 0 does equal 0. Therefore, we verify this.

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