1d the general solution of the differential $$ \frac{d y}{d x}=\sin x e^{\cos x} $$

Question

UpStudy AI · Accepted Answer

To solve the differential equation:

$$
\frac{dy}{dx} = \sin(x) e^{\cos(x)}
$$

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

$$
y = \int \sin(x) e^{\cos(x)} \, dx + C
$$

Let's evaluate the integral:

$$
\int \sin(x) e^{\cos(x)} \, dx
$$

Let’s use substitution:

1. **Let** $$ u = \cos(x) $$.
2. **Then**, $$ \frac{du}{dx} = -\sin(x) $$, which implies $$ -du = \sin(x) \, dx $$.

Substituting into the integral:

$$
\int \sin(x) e^{\cos(x)} \, dx = -\int e^{u} \, du = -e^{u} + C = -e^{\cos(x)} + C
$$

Therefore, the general solution to the differential equation is:

$$
y = -e^{\cos(x)} + C
$$

where $$ C $$ is an arbitrary constant.

**Final Answer:**

After simplifying, a general solution is –eᶜᵒˣ plus a constant. Thus,

y = − e^{cos x} + C
The general solution is $$ y = -e^{\cos(x)} + C $$.

1d the general solution of the differential \( \frac{d y}{d x}=\sin x e^{\cos x} \)