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anonymous

  • 5 years ago

d^2y/dx^2 - 2 dy/dx - 15y = 0 given y(0) = 10 y^1(0) = -7

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  1. anonymous
    • 5 years ago
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    well the differential equation is: \[y''-2y'-15y=0, y(0)=10, y'(0)=-7\] it can be easily solved using the auxiliary equations

  2. anonymous
    • 5 years ago
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    it's a 2nd order linear homogenous differential equation with constant co-efficents

  3. anonymous
    • 5 years ago
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    \[m^2-2m-15=0 \implies (m-5)(m+3)=0 \implies m=5, m=-3\] the general solution is: \[y(x)=c_1e ^{5x}+c_2e^{-3x}\] now we have just to solve for the initial conditions

  4. anonymous
    • 5 years ago
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    can you do the rest?

  5. anonymous
    • 5 years ago
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    I mean can you find c1 and c2? :)

  6. anonymous
    • 5 years ago
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    m^2 -2m -15 =0 (m+3)(m-5) gives you two real root solutions hence y=Ae^-3x + Be^5x no use the intial values to find A & B am i on the right path

  7. anonymous
    • 5 years ago
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    y= 57/8 e^-3x + 23/8 e^5x thats my answer

  8. anonymous
    • 5 years ago
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    \[y'(x)=5c_1e ^{5x}-3c_2e ^{-3x} \implies y'(0)=5c_1-3c_2=10 \rightarrow (1)\] no need to continue... you got it already

  9. anonymous
    • 5 years ago
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    your answer is perfectly right.. you still have one thing to do, which is to be a fan ;)

  10. anonymous
    • 5 years ago
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    Try www.aceyourcollegeclasses.com

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