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Math Help - First-Order Linear Differential Equation

  1. #1
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    First-Order Linear Differential Equation

    I am suppose to solve y'=5y; y(0)=1

    However, I don't understand how to do this as the standard form for this type of equation seems to be: y' + a(t)y = b(t)

    What do I use as b(t)?

    Everything I've done so far works out to zero. Could someone explain?
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  2. #2
    Super Member Random Variable's Avatar
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     y'-5y=0

    which has the characteristic equation  r-5=0

    so the general solution is  C_{1}e^{5t}

    but  y(0)=1=C_{1}

    so the final solution is  e^{5t}
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Quixotic View Post
    I am suppose to solve y'=5y; y(0)=1

    However, I don't understand how to do this as the standard form for this type of equation seems to be: y' + a(t)y = b(t)

    What do I use as b(t)?

    Everything I've done so far works out to zero. Could someone explain?

    y' = 5y

    Since this is separable,

    \int \frac{1}{5y} dy = \int 1 dx

    \frac{log(y)}{5} + C_1 = x + C_2

    \frac{log(y)}{5} = x + C_3 Note: C_3 = {C_2}-{C_1} is also a constant

    log(y) = 5x+ C_4 note: C_4 = 5 C_3 is a constant

    \therefore y = e^ {5x+{c_4}}

    y = e^{5x} \times e^{C_4}

    Thus,

    y= C e^{5x}.......(I) This is what Random Variable has gotten in the above post.

    Note:  C = e^{C_4} is also a constant.


    now use y(0)=1 [in(I)] to find C
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