$$ \left. \begin{array} { l } { y d x + ( y + an ( x + y ) ) d y = 0 } \ { d x } \end{array} ight. $$

Question

$$ \left. \begin{array} { l } { y d x + ( y + 	an ( x + y ) ) d y = 0 } \ { d x } \end{array} ight. $$

UpStudy AI · Accepted Answer

We start with the differential equation

$$
y\,dx + \Bigl(y+\tan(x+y)\Bigr)\,dy=0.
$$

A common trick for equations containing $$x+y$$ is to make the substitution

$$
u=x+y.
$$

Then we have

$$
du=dx+dy\quad\Longrightarrow\quad dx=du-dy.
$$

Substitute this into the original equation:

$$
y\,(du-dy) + \Bigl(y+\tan u\Bigr)\,dy=0.
$$

Distribute $$y$$:

$$
y\,du - y\,dy + y\,dy+\tan u\,dy=0.
$$

Notice that $$-y\,dy+y\,dy$$ cancel, leaving

$$
y\,du + \tan u\,dy=0.
$$

We can now rewrite the equation as

$$
y\,du = -\tan u\,dy.
$$

Assuming $$y\neq 0$$ and $$\tan u\neq 0$$, we separate variables by dividing both sides by $$y\,\tan u$$:

$$
\frac{du}{\tan u} = -\frac{dy}{y}.
$$

Recall that

$$
\frac{1}{\tan u}=\cot u.
$$

Thus the equation becomes

$$
\cot u\,du = -\frac{dy}{y}.
$$

Now we integrate both sides:

$$
\int \cot u\,du = -\int \frac{dy}{y}.
$$

Recall that the integral of $$\cot u$$ is

$$
\int \cot u\,du=\ln\left|\sin u\right|+C_1,
$$

and the integral on the right is

$$
\int \frac{dy}{y}=\ln\left|y\right|+C_2.
$$

Thus we have

$$
\ln\left|\sin u\right|=-\ln\left|y\right|+C,
$$

where $$C=C_1-C_2$$ is an arbitrary constant. One may rewrite the equation as

$$
\ln\left|\sin u\right|+\ln\left|y\right|=C.
$$

Using the logarithmic property $$\ln|a|+\ln|b|=\ln|ab|$$, we obtain

$$
\ln\left|y\sin u\right|=C.
$$

Exponentiating both sides gives

$$
\left|y\sin u\right|=e^C.
$$

Since $$e^C$$ is an arbitrary positive constant, we denote it by $$C_1$$. Dropping the absolute value and absorbing any sign into the constant, the general solution may be written as

$$
y\sin u=C,
$$

where $$C$$ is an arbitrary constant. Recalling that $$u=x+y$$, the final answer is

$$
y\sin(x+y)=C.
$$
The solution to the differential equation $$ y\,dx + (y + \tan(x + y))\,dy = 0 $$ is $$ y\sin(x + y) = C $$, where $$ C $$ is an arbitrary constant.

\( \left. \begin{array} { l } { y d x + ( y + \tan ( x + y ) ) d y = 0 } \\ { d x } \end{array} \right. \)

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