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Find the solution of the differential equation that satisfies the given initial condition.

$ y' \tan x = a + y, y(\pi/3) = a, 0 < x < \pi/2 $

$$y=\frac{4 a}{\sqrt{3}} \sin x-a$$

Differential Equations

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Campbell University

University of Michigan - Ann Arbor

University of Nottingham

Boston College

this question asked us to find the solution of the differential equation that satisfies the given initial condition. We know that we have why one tan of axe is a plus y. Now what we know we need to do is we need to get all the why terms on one side and oval ex terms on the other side. So what we know we have is, do you? Why, over a plus why is equivalent to co tangent of ex DX? We got coach tangent because we know that one over 10 is the same thing as one overcoat. Tension is the same thing as tent. So what we did was buy dividing by tangent. We know they could just consider that to essentially be co tangent when we switch this over to the right hand side. So it fits with the rest of the excess, not take the integral of book sides. And we know we end up with natural log of a plus. Y is natural log of sign of ax plus C or a constant of integration. No, we know we want to get a pulse y by itself. So we have a pulse. Y absolutely of this. Let's see. Sign of axe. Now, given the fact that excess power over three and wise A we know we have the absolute value of two. A is C square root of three over two, which means we now have four divided by squirt of three absolute value. A is C. Now we know the solution for this is a plus. Why is for a over squared of three sign of acts now? The goal, remember, is to always got this just in terms of a singular. Why? So we now have Why is for a over squirt of three sign of axe minus a