Solve the initial value problem.
-$$\frac { d y } { d x } = \csc x , y ( 1 ) = - 8$$
A) $y = - \int _ { 1 } ^ { x } \operatorname { csct } \cot t d t - 8$
B) $y = \int _ { - 8 } ^ { x } \operatorname { csctdt } + 1$
C) $y = \int _ { x } ^ { 1 } \operatorname { csctdt } - 8$
D) $y = \int _ { 1 } ^ { x } \operatorname { csctdt } - 8$

Question

Quizplus · Accepted Answer

Using the method of integrating factor, we multiply both sides by $\sin x$ to get
$$\sin x \frac{dy}{dx} = \frac{\sin x}{\sin x} \csc x = \frac{1}{\cos x}.$$
Noticing that the right-hand side is simply the derivative of $\sin x$, we integrate both sides to get
$$\sin x y = \int \frac{1}{\cos x} \, dx = \int \sec x \, dx = \ln |\sec x + \tan x| + C.$$
To find the value of $C$, we use the initial condition $y(1) = -8$, which gives
$$\sin 1 \cdot (-8) = \ln |\sec 1 + \tan 1| + C.$$
Simplifying this, we get
$$-8\sin 1 = \ln \left|\frac{1}{\cos 1} + \frac{\sin 1}{\cos 1}\right| + C = \ln |\tan (\frac{\pi}{4}+\frac{1}{2})| + C = \ln |1| + C = C,$$
where we have used the tangent half-angle formula and the fact that $\frac{\pi}{4} < 1 < \frac{\pi}{2}$. Thus our solution is
$$\sin x y = \ln |\sec x + \tan x| -8\sin 1,$$
which can be rewritten as
$$y = -\frac{1}{\sin x} \left(\ln |\frac{1}{\cos x} + \frac{\sin x}{\cos x}| -8\sin 1\right) = -\ln |\csc x + \cot x| + 8\cot 1.$$
Finally, we can verify that this matches the form given in choice $D$.

Solve the Initial Value Problem dydx=csc⁡x,y(1)=−8\frac { d y } { d x } = \csc x , y ( 1 ) = - 8dxdy​=cscx,y(1)=−8

Solve the Initial Value Problem $\frac { d y } { d x } = \csc x , y ( 1 ) = - 8$