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Math Help - Hard Dif. Equations

  1. #1
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    Hard Dif. Equations

    I have to solve the given Dif. EQs with the initial conditions that are given.

    But, the hard part is I'm not allowed to use the same method twice. (That is, I can only use separation of variables once, variation of parameters once, etc).

    1.) (1 + t^2)\frac{dy}{dt} = 2ty + 2, y(0) = -2

    2.) y''' - y'' - 2y' = e^{-t} + 3e^{2t}, y(0) = 1, y'(0) = \frac{-13}{6}, y''(0) = \frac{-5}{3}

    3.) \frac{dy}{dt} = \frac{t^2}{y + t^3y}, y(0) = -2

    4.) y'' + 4y' + 4y = t^{-2}e^{-2t}, y(1) = 0, y'(1) = 0
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    Quote Originally Posted by fifthrapiers View Post
    1.) (1 + t^2)\frac{dy}{dt} = 2ty + 2, y(0) = -2
    This is a linear ODE.

    Do you know how to do one of these?
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    Quote Originally Posted by fifthrapiers View Post
    3.) \frac{dy}{dt} = \frac{t^2}{y + t^3y}, y(0) = -2
    Factorise the denominator, then you'll be able to solve the separable ODE.
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    Quote Originally Posted by Krizalid View Post
    Factorise the denominator, then you'll be able to solve the separable ODE.
    Ahh, I definitely over looked this. Okay, does this look right:

    \displaystyle \frac{dy}{dt}=\frac{t^2}{y+t^3y}=\frac{t^2}{(t+1)(  t^2-t+1)y}

    \displaystyle y\cdot dy = \frac{t^2}{(t+1)(t^2-t+1)y}\cdot dt

    Take integral of both sides:

    \displaystyle \frac{y^2}{2} = \frac{\ln(|(t+1)(t^2-t+1)|)}{3} + C

    Use initial condition y(0) = -2

    \displaystyle \frac{(-2)^2}{2} = \frac{\ln(|(0+1)((0)^2-0+1)|)}{3} + C

    And so the particular solution, given the initial conditions is:

    \displaystyle \frac{y^2}{2} = \frac{\ln(|(t+1)(t^2-t+1)|)}{3} + 2
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    Quote Originally Posted by fifthrapiers View Post
    I have to solve the given Dif. EQs with the initial conditions that are given.

    But, the hard part is I'm not allowed to use the same method twice. (That is, I can only use separation of variables once, variation of parameters once, etc).

    1.) (1 + t^2)\frac{dy}{dt} = 2ty + 2, y(0) = -2

    2.) y''' - y'' - 2y' = e^{-t} + 3e^{2t}, y(0) = 1, y'(0) = \frac{-13}{6}, y''(0) = \frac{-5}{3}

    3.) \frac{dy}{dt} = \frac{t^2}{y + t^3y}, y(0) = -2

    4.) y'' + 4y' + 4y = t^{-2}e^{-2t}, y(1) = 0, y'(1) = 0
    I solved #2 using method of undetermined coefficients, so I don't need help on that anymore. Now I have #1 and #4 left (and if someone could check my #3). For #1, I think I have to use an integrating factor.....but not sure how to separate it into standard form first.. y'(1 + t^2) - 2 = 2ty ...

    That leaves me with #4... how can I solve this without using undetermined coefficients, separation of variables and integrating factor.. there's a method that is something involving constant coefficients, but not sure how to use that. I think I find y_c first and then y_p after... I can easily find y_c..its trying to find y_p that's hard.
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    Quote Originally Posted by fifthrapiers View Post
    4.) y'' + 4y' + 4y = t^{-2}e^{-2t}, y(1) = 0, y'(1) = 0
    No need to use variation of parameters use. Use the uniqueness theorem, notice that y = 0 for t>0 solves this equation trivial including the initial value problem. But the uniqueness theorem this is the only solution.
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    \left( {1 + t^2 } \right)y' - 2ty = 2 \iff y' - \frac{{2t}}<br />
{{1 + t^2 }}y = \frac{2}<br />
{{1 + t^2 }}

    So our I.F. will be \exp \left( { - \int {\frac{{2t}}<br />
{{1 + t^2 }}\,dt} } \right) = \left( {1 + t^2 } \right)^{ - 1}

    Multiply the equation by this I.F., then I assume you know you have to do.
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    Quote Originally Posted by Krizalid View Post
    \left( {1 + t^2 } \right)y' - 2ty = 2 \iff y' - \frac{{2t}}<br />
{{1 + t^2 }}y = \frac{2}<br />
{{1 + t^2 }}

    So our I.F. will be \exp \left( { - \int {\frac{{2t}}<br />
{{1 + t^2 }}\,dt} } \right) = \left( {1 + t^2 } \right)^{ - 1}

    Multiply the equation by this I.F., then I assume you know you have to do.
    Thanks a lot Kriz. I need some help finishing the integrating factor one:

    \left( {1 + t^2 } \right)y' - 2ty = 2 \iff y' - \frac{{2t}}{{1 + t^2 }}y = \frac{2}{{1 + t^2}}

    And so our integrating factor is:

    M(t) = e^{-\int{\frac{2t}{1 + t^2 }\,dt}} = e^{-\ln(x^2+1)} = \left( {1 + t^2} \right)^{-1}

    Multiply M(t) by both sides of the equation.

    \left( {1 + t^2 } \right)y'\underbrace{\left( {1 + t^2} \right)^{-1}} -\, 2ty\underbrace{\left( {1 + t^2} \right)^{-1}} = 2\underbrace{\left({1 + t^2}\right)^{-1}}

    Now this is where things get fuzzy. Is it just:

    \left[\left( {1 + t^2 } \right)y\left( {1 + t^2} \right)^{-1}\right]' = 2\left({1 + t^2}\right)^{-1}

    And then take integrals of both sides (which the integral of the derivative on the left will just leave what that is, so I just have to integrate the right side)...or is this step wrong?
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    Can anyone confirm if this is right or help me in the right direction?
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