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Sketch the direction field of the differential eq…

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Problem 13 Medium Difficulty

Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point.
$ y' = y + xy, (0,0) $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 9

Differential Equations

Section 2

Direction Fields and Euler's Method

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Video Transcript

We are given a differential equation and the points in the point we are asked on the slope field of this differential equation and to graph a solution which contains this differential equation. Given is y prime equals x, plus x y and the point in the plane is 01 in order to graph the slope field. Is that the given field number c, if y prime is equal to c a c is equal to x, plus x y? So this should actually be differential equation is y prime equals 6 tis equal to y plus x y is equal to y times 1 plus x for all x y such that the product will see, and so we have in particularly if x, is not equal to Negative 11 plus x is not equal to 0, and so y will be equal to c over 1 plus x and see that this is a translation of the reciprocal function. Y equals 1 if x, is equal to negative 1, and original ben tells us that y prime is equal to y minus y is equal to 0, so we have at the slope along the line x. Negative 1 is going to be 0 graphs function of the all the x and y axis and, as we are determined 1 x equals negative 1. We have that y prime is going to equal to zerothey, also have if we set c equal to 0. That y is equal to 0 x, not equal to negative 1 point and so along the x axis you a slope of 0. If c is equal to positive 1, you get that y is equal to 1 of the 1 plus x. This is simply function y equals at bex, shifting 1 metre to the left, so it has an asymptote x, equals negative 1 and x equals y equals 0, and the slope on these points will be c equal to 1 point. So we have a .01. As a point at negative .5 to and so on, we'll have a .1 fortini course. We have more slopes in between his long istate, also think at negative 2 negative 1 and at negative 1.5 negative 2 and so onwe see thet s c increases they're, taking the reciprocal function shifted over by 1 and we are essentially sketching it in the 1 direction And so, as you stretch for it and further to the x axis, the slopes are also very steeper. So approxinathis would be something like this fiction. We had more hyperboles with steeper and steeper slopes until we got to a point where they're essentially verticalyou can see that, if c is similar between 0 and 1, we have a contraction of function. Y equals 1 on the lumissowe'll have slopes there in between 0 and 1 limiteth points negative 10 and the curve y equals 1 of the ronxnowif c is sa negative 1 and y is equal to negative 11 plus x, which is going to be option. Y equals 111 plus x, put over the x axis and set each point is going to have a slope of negative 1 and particularly have 0 negative. 1 is a 1 to was negative, .5 negative 2 and in the second quadrant we have a similar thing on. We have negative 21 and the negative 1 and, as c gets more more negative or hyperboles get further and further away from the point negative 10, and it sleeps at epesthapproximation of this would be at the throne genitis as it pales get further and further. Like point slopes, its keeper and total almost vertical centuries c, is between 0 and negative 1 we're going to have slopes that between 0 and negative 1 mehan. So we get. It looks like a sort of dining pattern here where we seem to have most of the 0 slope. Points are concentrated around the asymptotes x equals negative 1 and y equals 0 point, and this is our slope field now to find solution containing 01 first 5.01. On this graph and as x, approaches infinity on the right of first will then see that we rapidly like an exponential function so that the graph approaches, positive infinity and as x, approaches negative. The graph decreases until it's question closer to the line y equals 0 with latin down point again, sort of like an exponent.

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Calculus: Early Transcendentals

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