$$ \left. \begin{array} { l } { \cos ( x + y ) d x + ( 3 y ^ { 2 } + 2 y + \cos ( x + y ) = 0 } \ { \frac { d y } { d x } + y = \sin x } \end{array} ight. $$

Question

$$ \left. \begin{array} { l } { \cos ( x + y ) d x + ( 3 y ^ { 2 } + 2 y + \cos ( x + y ) = 0 } \ { \frac { d y } { d x } + y = \sin x } \end{array} ight. $$

UpStudy AI · Accepted Answer

We are given the system
$$
\begin{cases}
\cos(x+y)\,dx + \left(3y^2+2y+\cos(x+y)\right)\,dy=0, \
\frac{dy}{dx}+y=\sin x.
\end{cases}
$$
We will solve each equation step by step.

---

### 1. Solving the First Differential Equation

The first equation is written in differential form:
$$
M(x,y)\,dx + N(x,y)\,dy=0,
$$
with
$$
M(x,y)=\cos(x+y) \quad\text{and}\quad N(x,y)=3y^2+2y+\cos(x+y).
$$

#### Checking for Exactness

A differential form is exact if
$$
\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}.
$$
We compute:
$$
\frac{\partial M}{\partial y}=\frac{\partial}{\partial y}\cos(x+y)=-\sin(x+y).
$$
Similarly,
$$
\frac{\partial N}{\partial x}=\frac{\partial}{\partial x}\left(3y^2+2y+\cos(x+y)\right)=-\sin(x+y).
$$
Since
$$
\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x},
$$
the equation is exact.

#### Finding the Potential Function

There exists a function $$ F(x,y) $$ such that
$$
F_x(x,y)=\cos(x+y) \quad \text{and} \quad F_y(x,y)=3y^2+2y+\cos(x+y).
$$

1. **Integrate $$ F_x(x,y)=\cos(x+y) $$ with respect to $$x$$:**

Treat $$y$$ as constant:
   $$
   F(x,y)=\int \cos(x+y)\,dx = \sin(x+y)+g(y),
   $$
   where $$g(y)$$ is an arbitrary function of $$y$$.

2. **Differentiate $$ F(x,y) $$ with respect to $$y$$:**

$$
   F_y(x,y)=\frac{\partial}{\partial y}\left[\sin(x+y)+g(y)\right] = \cos(x+y)+g'(y).
   $$

3. **Set $$ F_y(x,y) $$ equal to $$ N(x,y) $$:**

$$
   \cos(x+y)+g'(y)=3y^2+2y+\cos(x+y).
   $$
   Canceling the common term $$ \cos(x+y) $$ gives:
   $$
   g'(y)=3y^2+2y.
   $$

4. **Integrate to find $$g(y)$$:**

$$
   g(y)=\int (3y^2+2y)\,dy = y^3+y^2+C.
   $$
   We may absorb the constant $$ C $$ into the overall constant later.

Thus, the potential function is
$$
F(x,y)=\sin(x+y)+y^3+y^2.
$$

The exact equation implies
$$
F(x,y)=\text{constant},
$$
or equivalently,
$$
\sin(x+y)+y^3+y^2 = K,
$$
where $$ K $$ is an arbitrary constant.

---

### 2. Solving the Second Differential Equation

The second equation is a first–order linear differential equation:
$$
\frac{dy}{dx}+y=\sin x.
$$

#### Find the Integrating Factor

The standard form is:
$$
\frac{dy}{dx}+P(x)y=Q(x),
$$
with $$ P(x)=1 $$ and $$ Q(x)=\sin x $$. The integrating factor is
$$
\mu(x)=e^{\int P(x)\,dx}=e^{\int 1\,dx}=e^x.
$$

#### Multiply the Equation by $$\mu(x)$$

Multiplying by $$ e^x $$ gives:
$$
e^x\frac{dy}{dx}+e^xy = e^x\sin x.
$$
Notice that the left-hand side is the derivative of $$ e^xy $$:
$$
\frac{d}{dx}\bigl(e^xy\bigr)=e^x\sin x.
$$

#### Integrate Both Sides

Integrate with respect to $$ x $$:
$$
e^xy=\int e^x\sin x\,dx.
$$
Recall the standard integral:
$$
\int e^x\sin x\,dx=\frac{e^x}{2}\bigl(\sin x-\cos x\bigr)+C.
$$
Thus,
$$
e^xy=\frac{e^x}{2}\bigl(\sin x-\cos x\bigr)+C.
$$

#### Solve for $$ y $$

Divide both sides by $$ e^x $$:
$$
y=\frac{1}{2}\bigl(\sin x-\cos x\bigr)+Ce^{-x}.
$$

---

### Final Answers

1. The general solution of the exact differential equation is:
$$
\sin(x+y)+y^3+y^2=K.
$$

2. The general solution of the linear differential equation is:
$$
y=\frac{1}{2}\bigl(\sin x-\cos x\bigr)+Ce^{-x}.
$$
The solutions are: 1. $$ \sin(x + y) + y^3 + y^2 = K $$ 2. $$ y = \frac{1}{2}(\sin x - \cos x) + C e^{-x} $$

\( \left. \begin{array} { l } { \cos ( x + y ) d x + ( 3 y ^ { 2 } + 2 y + \cos ( x + y ) = 0 } \\ { \frac { d y } { d x } + y = \sin x } \end{array} \right. \)

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