Let $$y = y(x)$$ be the solution of the differential equation $$\frac{dy}{dx} = 2(y + 2\sin x - 5)x - 2\cos x$$ such that $$y(0) = 7$$. Then $$y(\pi)$$ is equal to

We start from the given differential equation

$$\frac{dy}{dx}=2\bigl(y+2\sin x-5\bigr)x-2\cos x.$$

The bracketed expression $$y+2\sin x-5$$ occurs as a whole, so it is natural to set

$$u=y+2\sin x-5.$$

Using the chain rule, the derivative of $$u$$ with respect to $$x$$ is

$$\frac{du}{dx}=\frac{dy}{dx}+2\cos x,$$

because $$\dfrac{d}{dx}\bigl(2\sin x\bigr)=2\cos x$$ and the derivative of the constant $$-5$$ is zero.

Now we substitute the given value of $$\dfrac{dy}{dx}$$ into this expression. From the differential equation we have

$$\frac{dy}{dx}=2x\bigl(y+2\sin x-5\bigr)-2\cos x=2xu-2\cos x.$$

Therefore

$$\frac{du}{dx}= \bigl(2xu-2\cos x\bigr)+2\cos x=2xu.$$

We have arrived at the much simpler separable equation

$$\frac{du}{dx}=2xu.$$

Re-arranging,

$$\frac{1}{u}\,du=2x\,dx.$$

We integrate both sides:

$$\int \frac{1}{u}\,du=\int 2x\,dx.$$

That gives

$$\ln|u|=x^{2}+C,$$

where $$C$$ is the constant of integration. Exponentiating both sides we get

$$u=C\,e^{x^{2}}.$$

Substituting back $$u=y+2\sin x-5$$, we obtain

$$y+2\sin x-5=C\,e^{x^{2}},$$

so

$$y=C\,e^{x^{2}}-2\sin x+5.$$

To evaluate the constant $$C$$ we use the initial condition $$y(0)=7$$. At $$x=0$$ we have $$\sin 0=0$$, hence

$$7=C\,e^{0}-2\sin 0+5=C\cdot1-0+5=C+5.$$

Therefore

$$C=7-5=2.$$

Thus the explicit solution is

$$y(x)=2\,e^{x^{2}}-2\sin x+5.$$

Now we need $$y(\pi)$$. Because $$\sin\pi=0$$, we get

$$y(\pi)=2\,e^{\pi^{2}}-2\cdot0+5=2e^{\pi^{2}}+5.$$

Hence, the correct answer is Option C.