82. Solve the equation $\frac{d y}{d x} \cos x+y \sin x=1$.
Solution. $1^{0}$. Dividing all terms of the equation by $\cos x$, we get

$$
\frac{d y}{d x}+y \operatorname{tg} x=\frac{1}{\cos x}
$$

$2^{0}$. Let $y=u v ; y^{\prime}=\frac{d y}{d x}=u^{\prime} v+v^{\prime} u$.

Substituting the expressions for $y$ and $\frac{d y}{d x}$ into the equation, we have

$$
u^{\prime} v+v^{\prime} u+u v \operatorname{tg} x=\frac{1}{\cos x}
$$

or

$$
u^{\prime} v+u\left(v^{\prime}+v \operatorname{tg} x\right)=\frac{1}{\cos x}
$$

$3^{0}$. Let $v^{\prime}+v \operatorname{tg} x=0$, or $\frac{d v}{d x}=-v \operatorname{tg} x$. Separating variables and integrating:

$$
\frac{d v}{v}=-\operatorname{tg} x d x ; \int \frac{d v}{v}=-\int \operatorname{tg} x d x \text { or } \ln v=\ln \cos x
$$

(we do not write the arbitrary constant $C$); hence $v=\cos x$.
$4^{0}$. Now equation (3) will take the form

$$
u^{\prime} \cos x=\frac{1}{\cos x}, \text { or } \frac{d u}{d x}=\frac{1}{\cos ^{2} x}, \text { or } d u=\frac{d x}{\cos ^{2} x} .
$$

Integrating, we find $\int d u=\int \frac{d x}{\cos ^{2} x}$, i.e., $u=\operatorname{tg} x+C$.

$5^{0}$. Substituting the values of $u$ and $v$ into the equality $y=u v$, we get $y=$ $=(\operatorname{tg} x+C) \cos x$, or, finally,

$$
y=\sin x+C \cos x
$$

83-92. Solve the equations: