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Show that each function $y=f(x)$ is a solution of the accompanying differential equation.$$y^{\prime}=y^{2}$$a. $y=-\frac{1}{x}$b. $y=-\frac{1}{x+3}$c. $y=-\frac{1}{x+C}$

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Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Integrals and Transcendental Functions

Section 2

Exponential Change and Separable Differential Equations

Functions

Trig Integrals

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We want to show that these three are solutions of our differential equation, so we could just take the derivative of each of those three functions, plug a man and go from there. But one thing that may help us cut down on a little bit of time is if we solve for C first and then use that to answer a and be released to show that it holds current. So let's start with C. So let's see. What? Why prime ISS or C? So we would write so negative deep I d. X of X plus seat raise the negative first so we would use power rule to take this derivative move the to outfront. Subtract one off of it. The negatives cancel out so that it's just going to be X plus C rays to the negative second, which we can reciprocate as one over X plus e to square. All right, so that's what Why pry Ms. Now let's see. What why squared is so why squared is Nega Do one x will see squared, which is just going to be so the negative goes away since we're squaring it and it speaks one over x plus e square. So why squared is equal to y fronts? Coursey, Right? So see checks out now for parts A and be actually. So for part, A A is the case. So it's heart. See when C is equal to zero so that one will check out as well. And B is part see when C is equal to three. So by just showing it holds for see first, it's Trivoli follows that A and B work as well.

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