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Slope Fields

Cosider the differential equation: dydx=f(x,y)\frac{dy}{dx}=f(x,y)

This differential gives a formula to compute the slope of any solution curve at any point (x,y)(x,y).

For Example: y′=x+yy'=x+y We can display the requirement that the tangent slope at every point on a solution curve be equal to x+yx+y with a slope field.

x,y=var('x,y')
f(x,y)=x+y
sf=plot_slope_field(f(x,y),(x,-3,3),(y,-3,3),aspect_ratio=1,figsize=8)
sf

We can solve the differential equation using a CAS:

y=function('y',x)
sol=desolve(diff(y,x)==x+y,y)
show(expand(desolve(diff(y,x)==x+y,y)))
Cex−x−1\displaystyle C e^{x} - x - 1

Every member of the family of curves y=Cex−x−1y=Ce^x-x-1 must be tangent to an element of the slope field at every point.

slns=sum([plot(C*e^x-x-1,(x,-3,3),ymin=-3,ymax=3) for C in srange(-3,3,.2)])
sf+slns

Slope Fields and IVPs

Consider the IVP: ParseError: KaTeX parse error: Undefined control sequence: \array at position 8: \left\{\̲a̲r̲r̲a̲y̲{\frac{dy}{dx}&…

The initial condition y(x0)=y0y(x_0)=y_0 specifies a particular solution among the family of curves that satisfy y′=f(x,y)y'=f(x,y).

Example

ParseError: KaTeX parse error: Undefined control sequence: \array at position 8: \left\{\̲a̲r̲r̲a̲y̲{\frac{dy}{dx}&…

The general solution to y′=x+yy'=x+y was already found to be the family of curves y=Cex−x−1y=Ce^x-x-1 (where CC is an arbitrary real number), and we can use the initial condition to specify CC:

ParseError: KaTeX parse error: Unknown column alignment: 0 at position 15: \begin{array} 0̲.75 &=& y(0.25)…

This value of CC specifies a particular member of the family of solutions to the ODE y′=x+yy'=x+y.

pnt=list_plot([(.25,.75)],color='red',size=100)
IVP_sln=plot((2/exp(.25))*e^x-x-1,(x,-3,3),ymin=-3,ymax=3,color='red')
IVP_sln+sf+slns+pnt