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For what values of $ r $ does the function $ y = e^{rx} $ satisfy the differential equation $ y" - 4y' + y = 0? $

$r=2 \pm \sqrt{3}$

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Okay, here's our function. Why equals E to the R X, and we're going to find out for what values of our it satisfies the differential equation given. So we need to find by prime. And we need to find why double prime and then substitute those into the differential equation. Okay, So using the chain rule, white crime will be each of the r x times, the derivative of Rx, which is our. So we could write this as our times e to the R X and then the second derivative we have the are the constant. And then we have the derivative of each of the Rx, which is just like we did in the first derivative. And we can write this as r squared each of the Rx. So now let's substitute are why double crime and are y prime into the differential equation and see what happens. So for white double prime, we have r squared each of the Rx, and then we have minus four times y prime. So minus four times are each of the r X plus why would be plus each of the Rx, and that equals zero Now, each of these terms has a factor of each of the Rx, so we can factor that out on the other factor is R squared minus four are plus one, so we can set each factor equal to zero. We get e to the R X equals zero and we get R squared. Minus four are plus one equals zero. Each of the Rx doesn't have any or each of the Rx equals one equals zero. Doesn't have any solutions because each willpower is always positive. So now let's work on the solutions to the other part. We can use the quadratic formula, and we get the opposite a B plus or minus the square root of B squared minus four a c over two a that gives us four plus or minus the square root of 12/2, and that gives us four plus or minus two square root 3/2 and all of those air divisible by two so we end up with R equals two plus or minus square root three

Oregon State University