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Maths

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Find the particular solution of the differential equation from:
dy/dx = 2/27(x-3)ã(x^2-6x+23) / (y) (y>0. For which y=2 when x =1, and then give this particular solution in explicit form.

regards Claire

  • Maths - ,

    Assuming ã stands for √, and assuming parentheses are as follows:

    dy/dx = (2/27)((x-3)√(x^2-6x+23)) / (y) (y>0. For which y=2 when x =1

    Separate variables:
    ydy = (2/27)(x-3)√(x^2-6x+23)
    complete squares and use substitution
    u=x-3
    du=dx
    ydy = (2/27)√(u^2+14) 2udu
    use substitution v=u^2+14, dv=2udu, integrate and backsubstitute:
    y^2/2 = (2(x^2-6x+23)^(3/2))/81 + C
    Substitute x=1, y=2 to find C, and hence y.
    I get C=2+2^(5/2).
    Check me.

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