Example 10. Solve the equation

$$
x \int_{0}^{x} y(t) d t=(x+1) \int_{0}^{x} t y(t) d t, \quad x>0
$$
Solution. Differentiating both sides of this equation with respect to $x$, we obtain

$$
\int_{0}^{x} y(t) d t+x y(x)=\int_{0}^{x} t y(t) d t+(x+1) x y(x)
$$

or

$$
\int_{0}^{x} y(t) d t=\int_{0}^{x} t y(t) d t+x^{2} y(x)
$$

Differentiating again with respect to $x$, we will have a linear homogeneous equation with respect to $y(x)$:

$$
y(x)=x y(x)+x^{2} y^{\prime}(x)+2 x y(x)
$$

or

$$
x^{2} y^{\prime}(x)+(3 x-1) y(x)=0
$$

Separating variables and integrating, we find

$$
y=C \frac{1}{x^{3}} e^{-1 / x}
$$

This solution, as is easily verified, satisfies the original equation.

## Problems for Independent Solution

Solve the following Bernoulli equations:

157. $y^{\prime}+2 x y=2 x y^{2}$
158. $\left(x^{3}+e^{y}\right) y^{\prime}=3 x^{2}$.
159. $y^{\prime}-2 y e^{x}=2 \sqrt{y} e^{x}$.
160. $2 y^{\prime} \sin x+y \cos x=y^{3} \sin ^{2} x$. 165. $y^{\prime}-y \cos x=y^{2} \cos x$.
161. $2 y^{\prime}-y=\frac{1}{y} e^{x}$.
162. $y^{\prime}+2 x y=y^{2} e^{x^{2}}$.
163. $2 y^{\prime} \ln x+\frac{y}{x}=y^{-1} \cos x$.
164. $\left(x^{2}+y^{2}+1\right) d y+x y d x=0$.

The following nonlinear equations can be reduced to linear or Bernoulli equations using a change of variables and solved:

166. $y^{\prime}-\operatorname{tg} y=e^{x} \frac{1}{\cos y}$.
167. $y^{\prime}=y\left(e^{x}+\ln y\right)$.
168. $y^{\prime} \cos y+\sin y=x+1$.
169. $y y^{\prime}+1=(x-1) e^{-y^{2} / 2}$.
170. $y^{\prime}+x \sin 2 y=2 x e^{-x^{2}} \cos ^{2} y$.

Solve the following equations using differentiation:
171. $\int_{0}^{x} t y(t) d t=x^{2} y(x)$.
172. $y(x)=\int_{0}^{x} y(t) d t+e^{x}$.
173. $\int_{a}^{x} t y(t) d t=x^{2}+y(x)$.
174. $\int_{0}^{1} y(x t) d t=n y(x)$.

## §7. Equations in Total Differentials. Integrating Factor

$1^{\circ}$. Equations in total differentials. A differential equation of the form

$$
M(x, y) d x+N(x, y) d y=0
$$

is called an equation in total differentials if its left-hand side represents the total differential of some function $u(x, y)$, i.e.,

$$
M d x+N d y \equiv d u \equiv \frac{\partial u}{\partial x} d x+\frac{\partial u}{\partial y} d y
$$

Theorem. For the equation (1) to be an equation in total differentials, it is necessary and sufficient that in some simply connected domain $D$ of variation of the variables $x$ and $y$, the condition

$$
\frac{\partial M}{\partial y} \equiv \frac{\partial N}{\partial x}
$$

is satisfied.

The general integral of equation (1) has the form $u(x, y)=C$ or

$$
\int_{x_{0}}^{x} M(x, y) d x+\int_{y_{0}}^{y} N(x, y) d y=C
$$