Wronskian

Wronskian

Lessons

A linear homogeneous second order differential equation is of the form:
a(x)y+b(x)y+c(x)y=0a(x) y''+b(x) y'+c(x)y=0

Let’s assume that we’ve found two solutions to the above differential equation, y1(x)=f(x)y_1 (x)=f(x) and y2(x)=g(x)y_2 (x)=g(x).

And the general solution will be of the form:
y(x)=c1f(x)+c2g(x)y(x)=c_1 f(x)+c_2 g(x)

We can us the Wronskian to see whether f(x)f(x) and g(x)g(x) are linearly independent

The Wronskian is defined as:

Wronskian

If we have two solution, f(x)f(x) and g(x)g(x), and W(f,g)0W(f,g)\neq0, then we say that f(x)f(x) and g(x)g(x) form a fundamental set of solutions, and the general solution will indeed be of the form:

y(x)=c1f(x)+c2g(x)y(x)=c_1 f(x)+c_2 g(x)
  • 1.
    How can we be sure that our general set of solutions is indeed the general set of solutions? The Wronskian!

  • 2.
    Verifying Some of Our General Solutions
    In the section “Characteristic Equation with Repeated Roots” we had the following differential equation:
    y6y+9y=0y''-6y'+9y=0

    It was found that the two solutions were:
    y1(x)=e3xy_1 (x)=e^{3x}
    y2(x)=xe3xy_2 (x)=xe^{3x}
    a)
    But one might have made the assumption that both solutions were of the form: y1(x)=y2(x)=e3xy_1 (x)=y_2 (x)=e^{3x}. Demonstrate that these solutions are not linearly independent

    b)
    Show that the two actual solutions actually form a fundamental set of solutions.


  • 3.
    In the section “Characteristic Equation with Real Distinct Roots” we had the following differential equation:
    6y+8y8y=06y''+8y'-8y=0
    We found two solutions:

    y1(x)=e23xy_1 (x)=e^{\frac{2}{3} x}
    y2(x)=e2xy_2 (x)=e^{-2x}

    Verify that these two solutions are indeed linearly independent

  • 4.
    A certain differential equation was found to have two solutions:
    y1(x)=3cos(2x)y_1(x)=3 \cos (2x)
    y2(x)=36sin2(x)y_2 (x)=3-6\sin^2 (x)
    Are these two solutions independent? Will they form a fundamental set of solutions?