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Denebel

  • 4 years ago

Solve the initial value problem. Confirm your answer by checking that it conforms to the slope field of the differential equation. dy/dx=(x+2)sin x and y=3 when x=0 I'm not sure what this problem is asking or how to solve it?

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  1. dumbcow
    • 4 years ago
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    integrate both sides \[y = \int\limits_{}^{}(x+2)\sin(x) dx = \int\limits_{}^{}x*\sin(x) +2\sin(x) dx\]

  2. malevolence19
    • 4 years ago
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    \[\frac{dy}{dx}=(x+2)\sin(x); y(0)=3\] \[\int\limits dy=\int\limits (x+2)\sin(x)dx\] Now it comes down to integration: \[y(x)=\int\limits x sin(x)dx+2 \int\limits \sin(x)dx\] Doing the first one using IBP you get: \[u=x; du=dx; dv=\sin(x)dx; v=-\cos(x)\] \[\int\limits x \sin(x)dx=-xcos(x)+\int\limits \cos(x)dx=-x \cos(x)+\sin(x)+C\] So we get: \[y(x)=-x \cos(x)+\sin(x)-2\cos(x)+C\] Solving the initial value we get: \[y(0)=3=-(0)\cos(0)+\sin(0)-2\cos(0)+C \implies 3=-2+C \implies C=5\] \[y(x)=-x \cos(x)+\sin(x)-2\cos(x)+5\]

  3. Denebel
    • 4 years ago
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    Okay I get it; thanks.

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