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Denebel
 3 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?
Denebel
 3 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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dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.0integrate both sides \[y = \int\limits_{}^{}(x+2)\sin(x) dx = \int\limits_{}^{}x*\sin(x) +2\sin(x) dx\]

malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{dy}{dx}=(x+2)\sin(x); y(0)=3\] \[\int\limits dy=\int\limits (x+2)\sin(x)dx\] 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: \[y(0)=3=(0)\cos(0)+\sin(0)2\cos(0)+C \implies 3=2+C \implies C=5\] \[y(x)=x \cos(x)+\sin(x)2\cos(x)+5\]
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