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- - **God help me (Differential Equations)**
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God help me (Differential Equations)I have this problem that is on our homework that is due Monday. We have had snow days and therefore we have not had our homework adjusted to the rate at which we are learning due to the snow closures. So as a result this problem is in out homework and I have NO clue how to do this. We have NOT even started second order differential equations and we are expected to have this done by tomorrow. Please help me. And thank you for your time in advance. Because this will take time. This is what you get when your math teacher has a Ph. D. in mathematics from Princeton. Lol http://img193.imageshack.us/img193/8209/imag0465.jpg Uploaded with ImageShack.us |

Re: God help me (Differential Equations)We are given: We can show, using the method of variation of parameters, that the expression given in the hint is a particular solution, to which we can then add the general solution to the corresponding homogeneous equation to get a general solution. First we find a fundamental solution set for the corresponding homogeneous equation and take: The auxiliary equation is and since we have a repeated root, we now have Next, we determine and by solving the system: First, we can divide both equations by since it can never be zero, to get: Solving the system, we find: Now, integrating to find and yields: Since we need only one particular solution, we take for simplicity. Now we substitute for and into the expression for to obtain a particular solution. Now, factoring from both terms gives: Or, written more efficiently using sigma notation: Recall that a general solution to a non-homogeneous equation is given by the sum of a general solution to the homogeneous equation and a particular solution. Consequently, a general solution to the given equation is: |

Re: God help me (Differential Equations)MarkFL, my hat tips to you. I am so thankful for your help and time!! I feel bad posting questions that I haven't taken a try at, but for this one, I was really unable to even start it. We have not had lectures on this material due to snow days and the homework packets have not been adjusted to the lecture schedule resulting in material being on the homework that we have not learned. I am so grateful for your help. So grateful. *hat tips* |

Re: God help me (Differential Equations)Yeah, I can't imagine being expected to solve such an equation without having had some lectures on the theory of linear 2nd order ODEs. It is possible there is a much simpler way to solve this equation, so check back as someone may know a more efficient way to go about it. :wink: |

e^(-2t)y'' - 4e^(-2t)y' + 4e^(-2t)y = 1 + t + t² + ... + t^29 e^(-2t)y' - 2e^(-2t)y = t + t²/2 + t³/3 + ... + t^30/30 + A e^(-2t)y = t²/2 + t³/(2*3) + ... + t^31/(30*31) + At + B y = (t²/2 + t³/(2*3) + ... + t^31/(30*31) + At + B)e^(2t), where A and B are constants. |

Re: God help me (Differential Equations)Ah, the "someone" I had in mind shows how it's done. 8) First, both sides were multiplied by to get: Then, the recognition that was used twice to write the equation as: Integrating twice: |

Re: God help me (Differential Equations)Quote:
had to be a better (i.e. faster) way! |

Re: God help me (Differential Equations)Quote:
[color=#008000]skipjack[/color]!In retrospect, a simple "check" of the solution would have easily shown me, but hindsight... :wink: |

I used the same method here, and you acknowledged you had read it. |

Re: God help me (Differential Equations)Yes, I remember now. I guess old habits die hard, but I will make an honest effort to remember this technique in the future, as it certainly is a time saver. Thank you for demonstrating a more efficient approach. |

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