Easiest way to draw Directional Field & Contour Curves | Geometry of 1st Differential Equations

In this video we are going to discuss geometrical meaning of first order differential equation which is also known as direction field and contour curves. Example dx by dt equal to 1. This is simplest first order differential equation where derivative is constant. So from the calculus we know that dx by dt is the slope of a curve xt. Hence if the above equation has a solution xt passing through t0 x0 it must have the slope 1 at t0 x0. It suggests the idea of plotting approximate solution of curve of the above equation. Suppose these are the points of solution curve xt. So dx by dt equal to 1. So at every solution point the slope is 1. So tan theta equal to 1. So theta equal We'll do 5x4. That means draw a tangent at the points of the solution curve XT with an angle pi by 4 with respect to T-axis. If this is the point of solution then we have to draw a tangent with the slope pi by 4. Now similarly if these are solution point then at every point slope is 1 therefore angle is pi by 4 of tangent. So by drawing the tangent with the angle pi by 4 This graph and this is called the directional field for the given differential equation. Now try to draw the solution curve xt such that at every point of the curve slope is 1. dx by dt equal to 1. As we know line is the only curve which has constant slope 1. So dx by dt equal to 1. So draw the curve xt at every point of the curve slope is 1. So this is a line passing through the solution point with the slope 1. Now we can draw the another line with the same slope 1 and we have infinitely many lines on the plane which has slope 1 and these are the solution for the differential equation dx by dt equal to 1 and these are called the contour curves x is equal to mt plus c equation of line where m equal to 1 so x is equal to t plus c is the solution of dx by dt equal to 1. Thank you friends. Now we generalize this concept. For the differential equations, suppose the form of differential equation is dy by dx is equal to f of xy. So, right hand side is not constant, but it is a function of xy. Then, step 1, draw the curve f of xy equal to k for some values of k. Step 2, indicate the slope dy by dx is equal to k as f of xy equal to k. So, dy by dx is also k. At some point of the curve, f of xy equal to k. By the short line segment on each of the above curves, it will give the directional field of the given differential equation. We repeat the process we discussed in the first example. Now step three in the directional field, sketch the approximate solution of the curve. This is called the contour curve that have the direction of the short line segment as their tangent direction. Example number two. Suppose dy by dx is equal to x. We are going to sketch the directional field and the contour curves. So dy by dx equal to x by compare with the standard form dy by dx is equal to fx. So fx is equal to x. Now we follow this step, step one and to draw f of xy equal to k for some values of k. So x is equal to k represent the vertical line. If k equal to 0 then f of xy equal to x equal to 0. This is the equation of y-axis. And the value of the slope on that curve is dy by dx is equal to f of xy equal to 0. So at k equal to 0, we have curve x is equal to 0. On the line x is equal to 0, every point has slope 0. Now k equal to 1, f of xy x is equal to 1. It's also represent the vertical line at the point 1, 0. And on that line, every point has a slope 1. Now, similarly, k equal to minus 1, then x is equal to minus 1, which is a vertical line passing through minus 1, 0 and the value of slope on that curve is minus 1. Now, if we combine this result, now similar for the other case, step 1. So, these are the points, solution points and these are the vertical lines. For x is equal to 0, x is equal to 1, x is equal to minus 1 and for the other values of x. Now for if k equal to 0 then f of x y equal to x equal to 0 is an equation of y axis and value of the slope at that curve is 0. So on this point on the solution point the slope is 0. Slope is 0 means the tangent is horizontal tangent. So just draw the horizontal tangent. Now for k equal to 1 slope is one on the line. x is equal to 1. Every solution point has the slope 1. So dy by dx is equal to 1. Now for x is equal to minus 1, every point has a slope minus 1. Now for the other values of k, we have this graph. This is a directional field for the differential equation dy by dx is equal to x. Now step 3, try to draw solution curve yx such that every point of the curve has a slope x dy by dx is equal to x. So we draw the approximate curve. So this curve is parabolic curve where at every point slope is dy by dx is equal to. Now for other curves, so we have infinitely many curves which has slope dy by dx is equal to x and these are the contour curves. Our solution is y is equal to x square by 2 plus c which are the solution of dy by dx is equal to x. Now these are the practice problems. Just draw the directional field for the dy by dx is equal to minus x by y which represent the circle. Another practice problem dy by dx is equal to x plus y. Thank you friends. Thanks for watching. If you learn from this video then like, subscribe and press the bell icon of next gen maths.