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 Differential Equations Ordinary and Partial Differential Equations Math Forum

September 5th, 2017, 11:56 AM   #1
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Solve (d^2)y/d(x^2) =y given that dy/dx =1 and y=1 when x=0.

I am not sure how to do this. What should the limits for v dv be?

Question 1 is the worked answer to the question. I am not sure where they get v=1, as it is not given in the question.
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Last edited by skipjack; September 5th, 2017 at 12:43 PM.

 September 5th, 2017, 01:22 PM #2 Global Moderator   Joined: Dec 2006 Posts: 17,908 Thanks: 1382 The image is difficult to read. When x = 0, v = dy/dx = 1. As $v = \dfrac{dy}{dx}$, $y = \dfrac{d^2y}{dx^2} = \dfrac{dv}{dx} = \dfrac{dv}{dy}\cdot\dfrac{dy}{dx} = v\dfrac{dv}{dy}$. Integrating $y = v\dfrac{dv}{dy}$ with respect to $y$ gives ${\small\dfrac12}v^2 = {\small\dfrac12}y^2 + c$, where $c$ is a constant. Note: it's asserted that $v = -y$ won't work, but that's incorrect. Is that sufficient help?

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