I'm so close to the answer, I've narrowed it down to two possibilites, here is the question: Find the initial value problem: Here are my two answers, only varying by the negative sign in front: or Any help is greatly appreciated!
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Originally Posted by KennyP33 I'm so close to the answer, I've narrowed it down to two possibilites, here is the question: Find the initial value problem: Here are my two answers, only varying by the negative sign in front: or Any help is greatly appreciated! Assuming that the IC is , it's the second answer.
Originally Posted by KennyP33 I'm so close to the answer, I've narrowed it down to two possibilites, here is the question: Find the initial value problem: Here are my two answers, only varying by the negative sign in front: or Any help is greatly appreciated! . This is first order linear, so use the integrating factor: . Multiplying through by the integrating factor gives: The left hand side is a product rule expansion of , so . So .
oh, wow, i forgot to add
Originally Posted by Prove It . This is first order linear, so use the integrating factor: . Multiplying through by the integrating factor gives: The left hand side is a product rule expansion of , so . So . I believe that and not what you have .
Originally Posted by KennyP33 oh, wow, i forgot to add Still - it's the second one.
so the one without the negative sign?
Originally Posted by KennyP33 so the one without the negative sign? Yep!
Originally Posted by Danny I believe that and not what you have . Thanks. So to fix up my mistake... The integrating factor is . So multiplying through we find . The LHS is a product rule expansion, so So . Now integrating by parts with so that and so that , we have . Now apply your initial conditions.
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