## Denebel Group Title 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? 2 years ago 2 years ago

1. dumbcow

integrate both sides $y = \int\limits_{}^{}(x+2)\sin(x) dx = \int\limits_{}^{}x*\sin(x) +2\sin(x) dx$

2. malevolence19

$\frac{dy}{dx}=(x+2)\sin(x); y(0)=3$ $\int\limits dy=\int\limits (x+2)\sin(x)dx$ 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

Okay I get it; thanks.