Solve the initial value problem.
-$$\frac { d y } { d x } = - 8 e ^ { - x } \sec e ^ { - x } \tan e ^ { - x } , y ( 0 ) = 8 \sec 1 + 8$$
A) $y = - 8 \sec x + 1$
B) $y = - 8 \sec e ^ { - x } + 1$
C) $y = 8 \sec e ^ { - x } + 8$
D) $y = 8 \tan e ^ { - x } + 8$

Question

Quizplus · Accepted Answer

We have:
$$\frac { d y } { d x } = - 8 e ^ { - x } \sec e ^ { - x } \tan e ^ { - x } $$

Separating the variables will give us:
$$\int \frac{1}{\sec(e^{-x})\tan(e^{-x})}dy=-\int 8e^{-x}dx$$
Now, using the substitution:
$$u=e^{-x},\hspace{5mm} \frac{du}{dx}=-e^{-x}$$
We can apply the substitution and we get:
$$-\int \frac{\cos u}{\sin u}du=\int 8du$$
$$-\ln|\sin u|=8x+C_1$$
$$-\ln|\sin e^{-x}|=8x+C_1$$
$$\ln|\frac{1}{\sin e^{-x}}|=-8x-C_1$$
Using the initial condition $y(0)=8\sec 1+8=\frac{8}{\cos 1}+8$
$$\ln|\frac{1}{\sin 1}|=-C_1\implies C_1=-\ln|\sin 1|$$

So, we end up with:
$$\ln|\frac{1}{\sin e^{-x}}|=-8x+\ln|\sin 1|$$

$$|\frac{1}{\sin e^{-x}}|=e^{-8x}e^{\ln|\sin 1|}=\frac{e^{-8x}}{\sin 1}$$

At this point, we can take the absolute value of the result again to remove the modulus and rearrange to get:
$$\frac{1}{y-8}= \sin e^{-x}$$

Now, we can subsitute $v=e^{-x}$ to get:
$$\frac{1}{y-8}=\sin v$$

$$\implies y-8=\frac{1}{\sin v}$$

$$\implies y-8= \csc e^{-x}$$

$$\implies y=8\sec e^{-x}+8$$

Therefore, the answer is choice C.

Solve the Initial Value Problem dydx=−8e−xsec⁡e−xtan⁡e−x,y(0)=8sec⁡1+8\frac { d y } { d x } = - 8 e ^ { - x } \sec e ^ { - x } \tan e ^ { - x } , y ( 0 ) = 8 \sec 1 + 8dxdy​=−8e−xsece−xtane−x,y(0)=8sec1+8

Solve the Initial Value Problem $\frac { d y } { d x } = - 8 e ^ { - x } \sec e ^ { - x } \tan e ^ { - x } , y ( 0 ) = 8 \sec 1 + 8$