Chapter 6.3, Problem 11E

To determine

To find: The initial equation.

Answer to Problem 11E

The initial equation: $y=-(x+2)\cos x+\sin x+4$

Explanation of Solution

Given information:

The differential equation is: $\frac{dy}{dx}=(x+2)\sin x$

And, $y=2$ when

  $x=0$

Calculation:

The given differential equation is: $\frac{dy}{dx}=(x+2)\sin x$

Integrate each side after separating the variables. Use part-by-part integration

  $\begin{array}{c}\frac{dy}{dx}=(x+2)\sin x\\ dy=(x+2)\sin x\cdot dx\\ \int dy=\int (x+2)\sin x\cdot dx\\ u=x+2\\ dv=\sin xdx\\ du=dx\\ v=-\cos x\\ y=(x+2)(-\cos x)-\int -\cos x\cdot dx\\ y=-(x+2)\cos x+\int \cos x\cdot dx\\ y=-(x+2)\cos x+\sin x+C\end{array}$

To find $C$ , use the initial conditions.

And, $y=2$ when

  $x=0$

  $\begin{array}{c}\text{}2=-(0+2)\cos 0+\sin 0+C\\ 2=-2+C\\ C=4\\ y=-(x+2)\cos x+\sin x+4\end{array}$

The solution is appropriate!

  

Therefore the required initial equation of given differential equation $\frac{dy}{dx}=(x+2)\sin x$ is $y=-(x+2)\cos x+\sin x+4$ .