Solve the initial value problem.
-$$\frac { d y } { d x } = \cot x , y ( - 2 ) = 9$$
A) $y = - \int _ { - 2 } ^ { x } \csc ^ { 2 } t d t + 9$
B) $y = \int _ { x } ^ { - 2 } \cot t d t + 9$
C) $y = \int _ { - 2 } ^ { x } \cot t d t + 9$
D) $y = \int _ { 9 } ^ { x } \cot t d t + - 2$

Question

Quizplus · Accepted Answer

We can solve the differential equation by integrating both sides with respect to x. However, since the integral of $\cot x$ is $\ln|\sin x|+C$, we cannot simply integrate $\cot x$ to get the solution. Instead, we use the fact that $\csc^2 x=1+\cot^2 x$ to rewrite the differential equation as $\frac{dy}{dx}=\frac{1}{\csc^2 x}=1+	an^2 x$. Now we can integrate both sides to get $$y=\int (1+	an^2 x)dx=x+\int 	an^2 x dx.$$ To evaluate the second integral, we use the identity $	an^2 x=\sec^2 x-1$ to get $$\int 	an^2 x dx=\int (\sec^2 x - 1) dx=	an x-x+C.$$ Therefore, the general solution to the differential equation is $$y=x+	an x-x+C=C+	an x.$$ To find the value of the constant C, we use the initial condition that $y(-2)=9$, which gives us $$9=C+	an(-2)=C-	an 2,$$ since $	an$ is an odd function. Therefore, $C=9+	an 2$, and the particular solution to the initial value problem is $$y=x+	an x+	an 2,$$ which is equivalent to choice C.

Solve the Initial Value Problem dydx=cot⁡x,y(−2)=9\frac { d y } { d x } = \cot x , y ( - 2 ) = 9dxdy​=cotx,y(−2)=9

Solve the Initial Value Problem $\frac { d y } { d x } = \cot x , y ( - 2 ) = 9$