Solve the differential equation.
$y ^ { t } = x ^ { 5 } e ^ { - \sin x } - y \cos x$
A) $y = \left( 6 x ^ { 6 } + C \right) e ^ { - x } \sin x$
B) $y = 5 C e ^ { - \sin x }$
C)
$y = \left( \frac { 1 } { 6 } x ^ { 6 } + C \right) e ^ { - \sin x }$
D) $y = e ^ { - \sin x } + 5 C x ^ { 5 }$
E) $y = \frac { 1 } { 6 } x + C e ^ { - \cos x }$

Question

Quizplus · Accepted Answer

Using the method of separation of variables, we get:

$$\frac{dy}{dx} = x^5e^{-\sin x} - y\cos x$$

Dividing both sides by $(x^5e^{-\sin x} - y\cos x)$, we get:

$$\frac{dy}{dx} \cdot \frac{1}{x^5e^{-\sin x} - y\cos x} = 1$$

Now, we can integrate both sides with respect to $x$:

$$\int \frac{1}{x^5e^{-\sin x} - y\cos x} \, dx = \int 1 \, dx$$

Let $u = \sin x$, then $du = \cos x \, dx$ and $x^5e^{-\sin x} = x^5e^{-u}$. Therefore, we can rewrite the left-hand side of the above equation as:

$$\int \frac{1}{x^5e^{-\sin x} - y\cos x} \, dx = -\int \frac{1}{x^5e^{-u} - y\cos \left(\sin^{-1} u\right)} \, du$$

We recognize the denominator as the derivative of the function $-e^{-u}$, so we can use the substitution $v = x^5e^{-u} - y\cos \left(\sin^{-1} u\right)$:

$$\int \frac{1}{x^5e^{-u} - y\cos \left(\sin^{-1} u\right)} \, du = -\int \frac{1}{5x^4 e^{-u}} \, dv = -\frac{1}{5x^4} \int e^v \, dv$$

$$=-\frac{1}{5x^4} e^{x^5e^{-\sin x} - y\cos x} + C$$

Therefore, the general solution to the differential equation is:

$$y = \left(\frac{1}{6} x^6 + C\right)e^{-\sin x}$$

which matches with option C.

Solve the Differential Equation yt=x5e−sin⁡x−ycos⁡xy ^ { t } = x ^ { 5 } e ^ { - \sin x } - y \cos xyt=x5e−sinx−ycosx

Solve the Differential Equation $y ^ { t } = x ^ { 5 } e ^ { - \sin x } - y \cos x$