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Mathematics 8 Online
OpenStudy (anonymous):

solve y'+3y=t using slope feild

OpenStudy (anonymous):

I hate to sound preachy, but solving a differential equation using a slope field is more something that you have to do (since its partly subjective), though I will guide you as to how to do it. Basically, either take some graph paper or draw a little mini version of your own. Then, choose a bunch of lattice points (ie. points with integer coordinates), particularly within the area around the origin and calculate the value of y' there. For example, if you choose (t,y)=(1,2) and plug this into your equation, you would get that y'=-5. You would then draw a short little line at (1,2) with slope -5. Repeating this for a number of points around the origin will give you a series of slopes at these points. This is your "slope field." Now comes the subjective part. Since differential equations always have a family of solutions, you have to pick several initial values corresponding to several curves in your solution family. For example, you first curve might have y(0)=0, another have y(0)=3, etc. Then, starting from this initial value, draw your curve by following the "flow" of the smaller lines you drew on your slope field. Repeat this three or so times and this will give you several solutions to your differential equation. See http://en.wikipedia.org/wiki/File:Slope_Field.png for an excellent diagram describing this process. If it helps, the actual solution to this differential equation is \[y(t)=Ae ^{-3t}+t/3-1/9\] Thus, your drawn curves should look approximately like this function for various values of A. Note that the value of A determines which solution you are choosing from your solution family. Feel free to ask if you need more help with this question :)

OpenStudy (anonymous):

That's not preachy, it's a very appropriate answer to a question like this one. These forums cannot and should not be "answer mines", and taking the time to explain how to come by a solution rather than just posting an answer is valuable and appreciated.

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