Interval of validity

Interval of validity

Lessons

Interval of Validity:
An interval of validity is the range in which a solution to our differential equation is valid. The equation must not have any ‘holes’ or ‘breaks’ in it and it must contain the initial condition.

We can use our initial condition to find what our exact solution is, and find out what is our interval of validity

Theorem:
For an Initial Value Problem (IVP) of the form:
y+p(x)y=g(x)y'+p(x)y=g(x)

With initial value: y(x0)=y0y(x_0 )=y_0
If both p(x) and g(x) are continuous functions on the interval α\alpha < xx < β\beta, with our initial condition x0x_0 being within this interval, then there exists a unique solution to the IVP above.
  • Introduction
    What is an interval of validity?

  • 1.
    Determining Intervals of Validity
    For each of the following differential equations, the corresponding slope field is provided.
    Sketch the solution for each differential equation with the specific initial conditions given. What is the interval of validity?
    a)
    dydx=1(x+1),y(0)=2 \frac{dy}{dx}=\frac{1}{(x+1)}, y(0)=2
    Determining the interval of validity with slope fields

    b)
    dydx=e2x2+2y,y(4)=4 \frac{dy}{dx}=e^{-2x^2}+2y, y(-4)=4
    Slope fields and interval of validity


  • 2.
    In the section to do with Bernoulli Equations the solution to the differential equation:

    dydx+1xy=y3\frac{dy}{dx}+\frac{1}{x} y=y^3

    with initial condition y(12)=1y(\frac{1}{2})=1 was found to be:

    y=12xy=\frac{1}{\sqrt{2x}}

    Plot out this solution and explicitly give its interval of validity.

  • 3.
    Solve the following differential equation:

    dydx=2xey\frac{dy}{dx}=-2xe^{-y}

    With initial conditions y(1)=0y(1)=0.

    What is the interval of validity for the solution?

  • 4.
    Solve the following differential equation:

    dydx=4xy2\frac{dy}{dx}=4xy^2
    a)
    What is the interval of validity with initial conditions y(0)=12y(0)=\frac{1}{2}?

    b)
    What is the interval of validity with initial conditions y(12)=2y(\frac{1}{2})=-2?