Orthogonal Trajectories
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Orthogonal Curves ....
Properties:
Remember that a family of curves can represent solutions in differential equations. Presented here below are curves that are tangent to the y-axis at the origin. x^2 + y^2 = 2cx the goal is to take derivatives with respect to x on both sides solving the result for 2c (yielding 2x + 2y(dy/dx) = 2c). The resulting equation will yield a term (at least one) with dy/dx. Now take the original equation and solve it in for 2c (yielding x + (y^2)/x = 2c)
Now, as shown above equate the two terms for 2c to each other, or 2x + 2y(dy/dx) = x + (y^2)/x, and solve this expression for dy/dx = (y^2 - x^2)/2yx.
"In summary we see that we can pass back and forth between a differential equation and its family of solution curves."(Krantz)
You can go to DE Ex2 here...D.E. Ex2.
You can go to DE Ex1 here ...D.E. Ex1.
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