Question

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?

Answer 1

integrate both sides

\[y = \int\limits_{}^{}(x+2)\sin(x) dx = \int\limits_{}^{}x*\sin(x) +2\sin(x) dx\]

Answer 2

\[\frac{dy}{dx}=(x+2)\sin(x); y(0)=3\]
\[\int\limits dy=\int\limits (x+2)\sin(x)dx\]
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:
\[y(0)=3=-(0)\cos(0)+\sin(0)-2\cos(0)+C \implies 3=-2+C \implies C=5\]
\[y(x)=-x \cos(x)+\sin(x)-2\cos(x)+5\]

Answer 3

Okay I get it; thanks.