Solve the initial value problems in Exercises 87 and 88.

Question

87. dy/dx = 1 + 1/x, y(1) = 3

Accepted Answer

Welcome back, everyone. Solve the initial value problem. D Y divided by DX is equal to 4 minus 2 divided by X and Y 2 equals 7. So for this problem, let's begin by multiplying both sides of our equation by DX. We're going to get the D Y equals or minus 2 divided by XD X. What we can do now is simply integrate both sides. So on the left hand side, the integral of the Y is equal to Y. And on the right-hand side, we want to integrate 4DX, which is 4 X, and we're going to subtract two multiplied by the integral of 1 divided by X, which is LN of the absolute value of X, right? And we're going to add a constant of integration C because we have an indefinite integral on the right-hand side. So now our goal is to identify the constant C. We can show that C is equal to Y minus 4 X plus 2 LM of the absolute value of X, right? And we're going to use the fact that Y is equal to 7, when X is equal to 2. So that C is equal to 7 minus 4 multiplied by 2 + 2 LN of 2. Now, let's go ahead and solve. C is equal to. We got 7 minus 8, that's -1, so we basically got 2 LN of 2 minus 1. So now that we have our constant, we can substitute into the equation of Y, right? To get our solution. So Y is going to be equal to 4 X minus 2 LN of the absolute value of X plus. To Alan of 2. -1 Additionally, we can now apply the properties of logarithms. So, let's go ahead and apply them to simplify the expression, we're going to get 4 X. Minus 2 lm of the absolute value of Now, here, what we're going to do is simply use the quotient control, right? So, we can factor out 2 and this basically gives us LN of 2 divided by X, right? So, if we're adding a negative sign, so let's show it step by step to clearly see what's happening. If we simply factor out negative 2. In parencies we're going to have LN of the absolute value of X minus LN of 2, right, then we're subtracting 1 and L LN of the absolute value of X minus LN 2 can be written as Ln of the absolute value of. X divided by 2 using the quotient troll so that our final answer can now be written as Y equals 4 X minus 2 ln of the absolute value of X divided by 2. -1. Let's label it and thank you for watching.

Solve the initial value problems in Exercises 87 and 88.87. dy/dx = 1 + 1/x, y(1) = 3

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Solve the initial value problems in Exercises 87 and 88.
87. dy/dx = 1 + 1/x, y(1) = 3