Show that the isoclines of y′ = x + y form a one-parameter family of lines

by RoRi

January 31, 2016

If $y$ is the solution of a differential equation, the points at which $y'$ has a constant value $C$ lie on a line for each $C$ . This line is called an isocline.

Show that the isoclines of the differential equation

$y' = x + y$

form a one-parameter family of straight lines. Make a plot of the isoclines corresponding to the slopes $0, \ \pm \frac{1}{2}, \ \pm 1, \ \pm \frac{3}{2}, \ \pm 2$ . Using these isoclines, construct a direction field and sketch the integral curve passing through the origin. Identify one of the integral curves is also an isocline.

The isoclines of $y' = x + y$ are the curves $x + y= C$ which implies $y = -x + C$ . These are straight lines with slope $-1$ . The integral curve passing through the origin is $y = -x$ . The isocline $x+y=-1$ is also an integral curve.

One comment

September 19, 2017 at 12:40 am

Mike says:

The integral curve passing through the origin is not y=-x. In that case y’=-1, which is not x+y=0 (as it should be if it is the solution of y’=x+y).

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Stumbling Robot

Show that the isoclines of y′ = x + y form a one-parameter family of lines

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