Solve the initial value problem.?

dy/dx = 6x^3, y(0) = 8

7 Answers

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  • 2 months ago

    dy/dx=6x^3, y(0)=8

    =>

    dy=6(x^3)dx

    =>

    y=6S(x^3)dx+C

    =>

    y=6[(x^4)/4]+C

    =>

    y=3(x^4)/2+C

    is the G.S.

    8=3(0)/2+C

    =>

    C=8

    =>

    y=3(x^4)/2+8

    is the particular solution.

  • 2 months ago

    dy/dx = 6x³

    y = 6.(1/4).x⁴ + k → where k in constant

    y = (3/2).x⁴ + k → given that: y(0) = 8

    (3/2).0⁴ + k = 8

    k = 89

    → y = (3/2).x⁴ + 8

  • Ian H
    Lv 7
    2 months ago

    Give each problem a try before posting.

    If stuck at least try to learn the method for next time

    y = (3/2)x^4 + C

    when x = 0 y = 8, so C = 8

    y = (3/2)x^4 + 8

  • rotchm
    Lv 7
    2 months ago

    You already asked many times such questions and were fully answered to you.

    Looks like that you are not trying at all. Is this the case? Are you just lazy or trolling?

    dy/dx = 6x^3

    Integrate both sides. What do you get?

    Answer that & we will take it from there. 

    Hopefully no one will spoil you the answer. That would be very irresponsible of them. 

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  • 2 months ago

    dy/dx = 6x^3 ---> dy = 6x^3 dx --->  y = 6/4  x^4 + C   apply IC

    y(0) = 8 --->  C = 8 --->   y = 3/2 x^4 + 8

  • 2 months ago

    Separate variables:

    dy = 6x^3

    Integrate:

    y = (3/2)(x^4) + C

    Plug in initial condition:

    8 = C

    y = (3/2)(x^4) + 8

  • Ash
    Lv 7
    2 months ago

    dy/dx = 6x³

    ∫dy = ∫ 6x³ dx

    y(x) = 6x⁴/4 + C

    y(x) = (3/2)x⁴ + C ......(1)

    Given y(0) = 8

    y(0) = (3/2)(0)⁴ + C

    8 = C

    y(x) = (3/2)x⁴ + 8

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