homogenous 2nd order diff equation with particular solutions?

A homogeneous second-order linear differential equation, two functions y1, y2, and a pair of

initial conditions are given. First verify that y1 and y2 are solutions of the differential equation. Then find a particular solution of the form y = c1y1 + c2y2 that satisfies the given initial


a) x^2(y'') +2xy' -6y =0, y1= x^2, y2=x^(-3),

y(2) =10, y'(2) = 15

2 Answers

  • ted s
    Lv 7
    3 weeks ago

    YOU certainly can replace y with x² or x^(-3) to show they satisfy the DE.

    then y = A x² + B x^(-3) means y(2) = 10 = 4 A + B / 8 and

    y '(2) = 15 = 2 A + 12 B / 32...2 equations in A & B ...solve

  • Vaman
    Lv 7
    3 weeks ago

    x^2 y'' +2 x y' -6y=0

    Let y1 and y2 be the solution

    x^2 y1''+2x y1' -6 y1=0 (1)

    x^2 y2''+ 2 xy2' -6y2=0 (2)

    Multiply 1 by y2 and 2 by y1 and subtract.

    x^2(y1'' y2 -y2'' y1) +2x(y1' y2-y1' y2)=0

    Define w(y1,y2)= Wronskian =y1' y2 - y2'y1

    If y1 = c y2, the y1 and y2 are dependent. In that case W=y1'y2-y2'y1=0

    x^2 dw/dx+2x w=0.

    d/dx(x^2 w)=0 Integrate it. We get w=k/x^2. k is a constant. Therefore, y1 and y2 are independent.

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