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  • 5 years ago

how to solve dy/dx=x+y,y(0)=1

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  1. anonymous
    • 5 years ago
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    Noting that this is a linear ordinary differential equation, we can rewrite it in the form \[\frac{dy}{dx} + p(x) \cdot y = g(x)\] where \[p(x) = -1, g(x) = x\]Finding the integration factor u(x) is trivial, as \[u(x) = e^{\int p(x) dx} = e^{-x}\] Multiplying both sides by the integration factor, we arrive at \[e^{-x} \frac{dy}{dx} + e^{-x} \cdot -y = e^{-x} \cdot x\] The left side may be rewritten as \[(e^{-x} \cdot y)' = e^{-x} \cdot x\] and integrating both sides, we find that \[\int (e^{-x} \cdot y)' dx = \int e^{-x} x dx\] which is \[e^{-x} \cdot y + c_1 = -e^{-x} (x+ 1) + c_2\] for some c_1, c_2. Letting \[c = c_2 - c_1\] we may rewrite the differential equation as \[e^{-x} y = -e^{-x} (x + 1) + c\] Solving for y, we arrive at \[y = \frac{-e^{-x} (x+ 1) + c}{e^{-x}} = c e^{x} - x - 1\] which is our general solution. Solving for c we find that \[y(0) = c \cdot e^{0} - 0 - 1 = c - 1 = 1\] thus c equals 2.

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