1. Решити диференцијалну једначину  $\left( {2xy + {x^2}y + \frac{{{y^3}}}{3}} \right)dx + \left( {{x^2} + {y^2}} \right)dy = 0$.

Покушајмо видети да ли је ово диференцијална једначина тоталног диференцијала.

$\underbrace {\left( {2xy + {x^2}y + \frac{{{y^3}}}{3}} \right)}_Pdx + \underbrace {\left( {{x^2} + {y^2}} \right)}_Qdy = 0$

$\frac{{\partial P}}{{\partial y}} = 2x + {x^2} + \frac{1}{3}3{y^2} = 2x + {x^2} + {y^2}$

$\frac{{\partial Q}}{{\partial x}} = 2x$

Добили смо да   $\frac{{\partial P}}{{\partial y}} \ne \frac{{\partial Q}}{{\partial x}}$,  па ово није једначина тоталног диференцијала.

Покушајмо пронаћи функцију  $\mu \left( {x,y} \right)$ којом ћемо помножити једначину тако да она постане једначина тоталног диференцијала, односно

$ \mu Pdx + \mu Qdy = 0$  тако да је  $\frac{{\partial \left( {\mu P} \right)}}{{\partial y}} = \frac{{\partial \left( {\mu Q} \right)}}{{\partial x}}$

 

Како је

$\frac{1}{P}\left( {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} \right) = \frac{1}{{2xy + {x^2}y + \frac{{{y^3}}}{3}}} \cdot \left( {{x^2} + {y^2}} \right)$  функција која зависи од обе променљиве

$\frac{1}{Q}\left( {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} \right) = \frac{1}{{{x^2} + {y^2}}} \cdot \left( {{x^2} + {y^2}} \right) = 1$  константа        $    \Rightarrow    \mu  = \mu \left( x \right)$

Пошто је  $ \mu$  функција која зависи само од  $x$  важи да је  $\frac{{\partial \mu }}{{\partial y}} = 0$  и  $\frac{{\partial \mu }}{{\partial x}} = \mu '$

 

$\frac{{\partial \left( {\mu P} \right)}}{{\partial y}} = \frac{{\partial \left( {\mu Q} \right)}}{{\partial x}}$

$\underbrace {\frac{{\partial \mu }}{{\partial y}}}_{ = 0}P + \mu \frac{{\partial P}}{{\partial y}} = \underbrace {\frac{{\partial \mu }}{{\partial x}}}_{ = \mu '}Q + \mu \frac{{\partial Q}}{{\partial x}}$

$\mu \left( {2x + {x^2} + {y^2}} \right) = \mu '\left( {{x^2} + {y^2}} \right) + \mu  \cdot 2x$

$\mu \left( {2x + {x^2} + {y^2} - 2x} \right) = \mu '\left( {{x^2} + {y^2}} \right)$

$\mu \left( {{x^2} + {y^2}} \right) = \mu '\left( {{x^2} + {y^2}} \right)$

$\mu  = \mu '$

$\mu  = \frac{{d\mu }}{{dx}}$

$dx = \frac{{d\mu }}{\mu }{\rm{   }} {\text{         }}{\text{         }}  /\int {} $

$x = \ln \mu $

$\mu  = {e^x}$

 

Сада када смо нашли функцију  $\mu  $  можем,о прећи на решавање једначине тоталног диференцијала

\[\underbrace {{e^x}\left( {2xy + {x^2}y + \frac{{{y^3}}}{3}} \right)}_{{P_1}}dx + \underbrace {{e^x}\left( {{x^2} + {y^2}} \right)}_{{Q_1}}dy = 0\]

$\frac{{\partial {P_1}}}{{\partial y}} = \underbrace {\left( {{e^x}} \right)_y^,}_{ = 0}\left( {2xy + {x^2}y + \frac{{{y^3}}}{3}} \right) + {e^x}\left( {2xy + {x^2}y + \frac{{{y^3}}}{3}} \right)_y^, = $

${\text{         }}{\text{         }}  {\text{         }}{\text{         }}  {\text{         }}{\text{         }}  = {e^x}\left( {2x + {x^2} + \frac{1}{3}3{y^2}} \right) = {e^x}\left( {2x + {x^2} + {y^2}} \right)$

$\frac{{\partial {Q_1}}}{{\partial x}} = \left( {{e^x}} \right)_x^,\left( {{x^2} + {y^2}} \right) + {e^x}\left( {{x^2} + {y^2}} \right)_x^, = {e^x}\left( {{x^2} + {y^2}} \right) + {e^x} \cdot 2x = $

$ {\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}   = {e^x}\left( {{x^2} + {y^2} + 2x} \right)$

$\frac{{\partial {P_1}}}{{\partial y}} = \frac{{\partial {Q_1}}}{{\partial x}}$

$\exists F\left( {x,y} \right) = C:\underbrace {\frac{{\partial F}}{{\partial x}}}_{{P_1}}dx + \underbrace {\frac{{\partial F}}{{\partial y}}}_{{Q_1}}dy = 0$

 

$\frac{{\partial F}}{{\partial x}} = {P_1} = {e^x}\left( {2xy + {x^2}y + \frac{{{y^3}}}{3}} \right){\rm{   }}{\text{         }}{\text{         }}  /\int {} dx$

$\int {\frac{{\partial F}}{{\partial x}}dx = \int {{e^x}\left( {2xy + {x^2}y + \frac{{{y^3}}}{3}} \right)} } dx$ 

$ {\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }} {\text{         }}{\text{         }}    \left| {\begin{array}{*{20}{c}}
{2xy + {x^2}y + \frac{{{y^3}}}{3} = u}&{{e^x}dx = dv}\\
{\left( {2y + 2xy} \right)dx = du}&{v = \int {{e^x}dx = {e^x}} }
\end{array}} \right| $

$F\left( {x,y} \right)= \left( {2xy + {x^2}y + \frac{{{y^3}}}{3}} \right){e^x} - \int {{e^x}\left( {2y + 2xy} \right)dx}  $

$ {\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }} {\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }} {\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}  \left| {\begin{array}{*{20}{c}}
{2y + 2xy = u}&{{e^x}dx = dv}\\
{2ydx = du}&{v = {e^x}}
\end{array}} \right|  $

$ {\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }} {\text{         }}{\text{         }}    = \left( {2xy + {x^2}y + \frac{{{y^3}}}{3}} \right){e^x} - \left( {\left( {2y + 2xy} \right){e^x} - \int {{e^x}} 2ydx} \right)  $

$ {\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }} {\text{         }}{\text{         }}{\text{         }}{\text{         }}   {\text{         }}{\text{         }}    = \left( {2xy + {x^2}y + \frac{{{y^3}}}{3}} \right){e^x} - \left( {2y + 2xy} \right){e^x} + 2y{e^x} + \varphi \left( y \right) = $

$ {\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }}  {\text{         }}{\text{         }}{\text{         }}{\text{         }} {\text{         }}{\text{         }}    = \left( {2xy + {x^2}y + \frac{{{y^3}}}{3} - 2y - 2xy + 2y} \right){e^x} + \varphi \left( y \right) = $

$F\left( {x,y} \right) = \left( {{x^2}y + \frac{{{y^3}}}{3}} \right){e^x} + \varphi \left( y \right){\rm{   }}{\text{         }}{\text{         }}  /\frac{\partial }{{\partial y}}$

$\frac{{\partial F}}{{\partial y}} = \left( {{x^2} + {y^2}} \right){e^x} + \varphi '\left( y \right) = {Q_1}$

$\left( {{x^2} + {y^2}} \right){e^x} + \varphi '\left( y \right) = {e^x}\left( {{x^2} + {y^2}} \right)$

$\varphi '\left( y \right) = 0$

$\varphi \left( y \right) = {C_1}$

 

\[F\left( {x,y} \right) = \left( {{x^2}y + \frac{{{y^3}}}{3}} \right){e^x} + {C_1}\]

\[\left( {{x^2}y + \frac{{{y^3}}}{3}} \right){e^x} = C\]

 

2. Решити диференцијалну једначину  $\frac{y}{x}dx + \left( {{y^3} - \ln x} \right)dy = 0$.

Покушајмо видети да ли је ово диференцијална једначина тоталног диференцијала.

$\underbrace {\frac{y}{x}}_Pdx + \underbrace {\left( {{y^3} - \ln x} \right)}_Qdy = 0$

$\frac{{\partial P}}{{\partial y}} = \frac{1}{x}$

$\frac{{\partial Q}}{{\partial x}} =  - \frac{1}{x}$

Добили смо да   $\frac{{\partial P}}{{\partial y}} \ne \frac{{\partial Q}}{{\partial x}}$,  па ово није једначина тоталног диференцијала.

Покушајмо пронаћи функцију  $\mu \left( {x,y} \right)$ којом ћемо помножити једначину тако да она постане једначина тоталног диференцијала, односно

$ \mu Pdx + \mu Qdy = 0$  тако да је  $\frac{{\partial \left( {\mu P} \right)}}{{\partial y}} = \frac{{\partial \left( {\mu Q} \right)}}{{\partial x}}$

 

Како је

$\frac{1}{P}\left( {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} \right) = \frac{x}{y}\left( {\frac{1}{x} - \left( { - \frac{1}{x}} \right)} \right) = \frac{x}{y}\frac{2}{x} = \frac{2}{y} {\text{         }}{\text{         }}  \Rightarrow{\text{         }}{\text{         }}  \mu  = \mu \left( y \right)$

Пошто је  $ \mu$  функција која зависи само од  $y$  важи да је  $\frac{{\partial \mu }}{{\partial x}} = 0$  и  $\frac{{\partial \mu }}{{\partial y}} = \frac{{d\mu }}{{dy}} = \mu '$

 

$\frac{{\partial \left( {\mu P} \right)}}{{\partial y}} = \frac{{\partial \left( {\mu Q} \right)}}{{\partial x}}$

$\underbrace {\frac{{\partial \mu }}{{\partial y}}}_{ = \mu '}P + \frac{{\partial P}}{{\partial y}}\mu  = \underbrace {\frac{{\partial \mu }}{{\partial x}}}_{ = 0}Q + \frac{{\partial Q}}{{\partial x}}\mu $

$\mu '\frac{y}{x} + \frac{1}{x}\mu  =  - \frac{1}{x}\mu $

$\mu '\frac{y}{x} =  - \frac{2}{x}\mu $

$\frac{{d\mu }}{{dy}}y =  - 2\mu $

$\frac{{d\mu }}{\mu } =  - 2\frac{{dy}}{y}{\rm{   }{\text{         }}{\text{         }}  }/\int {} $

$\ln \mu  =  - 2\ln y$

$\mu  = \frac{1}{{{y^2}}}$

 

Сада када смо нашли функцију  $\mu  $  можем,о прећи на решавање једначине тоталног диференцијала

\[\frac{1}{{{y^2}}}\frac{y}{x}dx + \frac{1}{{{y^2}}}\left( {{y^3} - \ln x} \right)dy = 0\]

\[\underbrace {\frac{1}{{xy}}}_{{P_1}}dx + \underbrace {\left( {y - \frac{{\ln x}}{{{y^2}}}} \right)}_{{Q_1}}dy = 0\]

$\frac{{\partial {P_1}}}{{\partial y}} = \frac{1}{x}\left( { - 1{y^{ - 2}}} \right) =  - \frac{1}{{x{y^2}}}$

$\frac{{\partial {Q_1}}}{{\partial x}} =  - \frac{1}{{{y^2}}}\frac{1}{x} = \frac{1}{{x{y^2}}}$

$\frac{{\partial {P_1}}}{{\partial y}} = \frac{{\partial {Q_1}}}{{\partial x}}$

$\exists F\left( {x,y} \right) = C:\underbrace {\frac{{\partial F}}{{\partial x}}}_{{P_1}}dx + \underbrace {\frac{{\partial F}}{{\partial y}}}_{{Q_1}}dy = 0$

 

$\frac{{\partial F}}{{\partial x}} = {P_1} = \frac{1}{{xy}}{\rm{   }}{\text{         }}{\text{         }}  /\int {} dx$

$F\left( {x,y} \right) = \int {\frac{1}{{xy}}} dx$

$F\left( {x,y} \right) = \frac{1}{y}\ln x + \varphi \left( y \right){\rm{   }} {\text{         }}{\text{         }}  /\frac{\partial }{{\partial y}}$

$\frac{{\partial F}}{{\partial y}} = \ln x \cdot \left( { - {y^{ - 2}}} \right) + \varphi '\left( y \right) = {Q_1}$

$ - \frac{{\ln x}}{{{y^2}}} + \varphi '\left( y \right) = y - \frac{{\ln x}}{{{y^2}}}$

$\varphi '\left( y \right) = y{\rm{   }} {\text{         }}{\text{         }}  /\int {} dy$

$\varphi \left( y \right) = {y^2} + C$

\[F\left( {x,y} \right) = \frac{{\ln x}}{y} + {y^2} + {C_1}\]

\[\frac{{\ln x}}{y} + {y^2} = C\]

 

 

3. Решити диференцијалну једначину  $xdx + ydy + xdy - ydx = 0$,  ако знамо да има интеграциони множитељ облика  $h = h\left( {{x^2} + {y^2}} \right)$.

$xdx + ydy + xdy - ydx = 0$

$\underbrace {\left( {x - y} \right)}_Pdx + \underbrace {\left( {x + y} \right)}_Qdy = 0$

$\frac{{\partial P}}{{\partial y}} =  - 1$

$\frac{{\partial Q}}{{\partial x}} = 1$

$\frac{{\partial P}}{{\partial y}} \ne \frac{{\partial Q}}{{\partial x}}$

Ово није једначина тоталног диференцијала, па тражимо функцију  $h = h\left( {{x^2} + {y^2}} \right)$  којом када помножимо једначину добијамо једначину тоталног диференцијала.

$\underbrace {h \cdot \left( {x - y} \right)}_{{P_1}}dx + \underbrace {h \cdot \left( {x + y} \right)}_{{Q_1}}dy = 0$

односно тако да важи

$\frac{{\partial {P_1}}}{{\partial y}} = \frac{{\partial {Q_1}}}{{\partial x}}$

$\frac{{\partial \left( {h \cdot \left( {x - y} \right)} \right)}}{{\partial y}} = \frac{{\partial \left( {h \cdot \left( {x + y} \right)} \right)}}{{\partial x}}$

$\frac{{\partial h}}{{\partial y}} \cdot \left( {x - y} \right) + \frac{{\partial \left( {x - y} \right)}}{{\partial y}} \cdot h = \frac{{\partial h}}{{\partial x}} \cdot \left( {x + y} \right) + \frac{{\partial \left( {x + y} \right)}}{{\partial x}} \cdot h$

 

Како је  $h = h\left( t \right)$  уводимо смену  ${x^2} + {y^2} = t$.

Тада је

$h = h\left( t \right)$

$\frac{{\partial h}}{{\partial x}} = \frac{{\partial h}}{{\partial t}}\frac{{\partial t}}{{\partial x}} = h' \cdot 2x$

$\frac{{\partial h}}{{\partial y}} = \frac{{\partial h}}{{\partial t}}\frac{{\partial t}}{{\partial y}} = h' \cdot 2y$

 

$h' \cdot 2y \cdot \left( {x - y} \right) + \left( { - 1} \right) \cdot h = h' \cdot 2x \cdot \left( {x + y} \right) + 1 \cdot h$

$h' \cdot \left( {2xy - 2{y^2} - 2{x^2} - 2xy} \right) = 2h$

$h'\left( { - 2\underbrace {\left( {{x^2} + {y^2}} \right)}_{ = t}} \right) = 2h$

$ - h't = h$

$ - \frac{{dh}}{{dt}}t = h$

$\frac{{dh}}{h} =  - \frac{{dt}}{t}{\rm{   }}{\text{         }}{\text{         }}  /\int {} $

$\ln h =  - \ln t$

$h = \frac{1}{t}$

$h = \frac{1}{{{x^2} + {y^2}}}$

 

Сада, када смо добили функцију  $h$  можемо прећи на решавање једначине тоталног диференцијала

\[\underbrace {\frac{{x - y}}{{{x^2} + {y^2}}}}_{{P_1}}dx + \underbrace {\frac{{x + y}}{{{x^2} + {y^2}}}}_{{Q_1}}dy = 0\]

$\frac{{\partial {P_1}}}{{\partial y}} = \frac{{ - 1 \cdot \left( {{x^2} + {y^2}} \right) - \left( {x - y} \right) \cdot 2y}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} = \frac{{ - {x^2} - {y^2} - 2xy + 2{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} = \frac{{ - {x^2} + {y^2} - 2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$

$\frac{{\partial {Q_1}}}{{\partial x}} = \frac{{1 \cdot \left( {{x^2} + {y^2}} \right) - \left( {x + y} \right) \cdot 2x}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} = \frac{{{x^2} + {y^2} - 2{x^2} - 2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} = \frac{{ - {x^2} + {y^2} - 2xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$

$\frac{{\partial {P_1}}}{{\partial y}} = \frac{{\partial {Q_1}}}{{\partial x}}$

$\exists F\left( {x,y} \right) = C:\underbrace {\frac{{\partial F}}{{\partial x}}}_{{P_1}}dx + \underbrace {\frac{{\partial F}}{{\partial y}}}_{{Q_1}}dy = 0$

 

$\frac{{\partial F}}{{\partial x}} = {P_1} = \frac{{x - y}}{{{x^2} + {y^2}}}{\rm{   }} {\text{         }}{\text{         }}  /\int {} dx$

$\int {\frac{{\partial F}}{{\partial x}}dx = \int {\frac{{x - y}}{{{x^2} + {y^2}}}dx} } $

$F\left( {x,y} \right) = \int {\frac{x}{{{x^2} + {y^2}}}dx}  - \int {\frac{y}{{{x^2} + {y^2}}}dx} $

${\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }}  \left| {\begin{array}{*{20}{c}}
{{x^2} + {y^2} = t}\\
{2xdx = dt}\\
{xdx = \frac{{dt}}{2}}
\end{array}} \right|$

${\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }}    = \int {\frac{{\frac{{dt}}{2}}}{t}}  - y\int {\frac{1}{{{x^2} + {y^2}}}dx} $

${\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }}    = \frac{1}{2}\ln \left| {{x^2} + {y^2}} \right| - y\frac{1}{y}arctg\frac{x}{y} + \varphi \left( y \right)$

$F\left( {x,y} \right) = \frac{1}{2}\ln \left| {{x^2} + {y^2}} \right| - arctg\frac{x}{y} + \varphi \left( y \right)$

Да бисмо још одредили  $\varphi \left( y \right)$,  извршићемо парцијалну интеграцију функције  $F$  по  $y$  и то изједначити са  $Q_1$.

$F\left( {x,y} \right) = \frac{1}{2}\ln \left| {{x^2} + {y^2}} \right| - arctg\frac{x}{y} + \varphi \left( y \right){\rm{   }}/\frac{\partial }{{\partial y}}$

$\frac{{\partial F}}{{\partial y}} = \frac{1}{2} \cdot \frac{1}{{{x^2} + {y^2}}} \cdot 2y - \frac{1}{{1 + {{\left( {\frac{x}{y}} \right)}^2}}} \cdot x\left( { - {y^{ - 2}}} \right) + \varphi '\left( y \right) = {Q_1}$

$\frac{y}{{{x^2} + {y^2}}} - \frac{{{y^2}}}{{{x^2} + {y^2}}}\left( { - \frac{x}{{{y^2}}}} \right) + \varphi '\left( y \right) = {Q_1}$

$\frac{{y + x}}{{{x^2} + {y^2}}} + \varphi '\left( y \right) = {Q_1}{\rm{   }} \Rightarrow {\rm{   }}\varphi '\left( y \right) = 0{\rm{   }} \Rightarrow {\rm{   }}\varphi \left( y \right) = C$

\[F\left( {x,y} \right) = \frac{1}{2}\ln \left| {{x^2} + {y^2}} \right| - arctg\frac{x}{y} + C\]

\[\frac{1}{2}\ln \left| {{x^2} + {y^2}} \right| - arctg\frac{x}{y} = C\]

 

 

4. Решити диференцијалну једначину  ${y^2}dx + {x^2}dy = 0$   ако се зна да има интеграциони множитељ облика  $h = h\left( {x + y} \right)$.

\[\underbrace {{y^2}}_Pdx + \underbrace {{x^2}}_Qdy = 0\]

$\frac{{\partial P}}{{\partial y}} = 2y$

$\frac{{\partial Q}}{{\partial x}} = 2x$

$\frac{{\partial P}}{{\partial y}} \ne \frac{{\partial Q}}{{\partial x}}$

$h = h\left( {x + y} \right):{\rm{   }}hPdx + hQdy = 0$

${\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }}   {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }}    {\text{         }}{\text{         }} {\text{         }}{\text{         }}     \frac{{\partial hP}}{{\partial y}} = \frac{{\partial hQ}}{{\partial x}}$

${\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }}    \frac{{\partial h}}{{\partial y}}P + \frac{{\partial P}}{{\partial y}}h = \frac{{\partial h}}{{\partial x}}Q + \frac{{\partial Q}}{{\partial x}}h$

 

$x + y = t {\text{         }}{\text{         }}     \Rightarrow{\text{         }}{\text{         }}      h = h\left( t \right)$

$\frac{{\partial h}}{{\partial y}} = \frac{{\partial h}}{{\partial t}} \cdot \frac{{\partial t}}{{\partial y}} = h' \cdot 1 = h'$

$\frac{{\partial h}}{{\partial x}} = \frac{{\partial h}}{{\partial t}} \cdot \frac{{\partial t}}{{\partial x}} = h' \cdot 1 = h'$

 

$h'{y^2} + 2yh = h'{x^2} + 2xh$

$h'\left( {{y^2} - {x^2}} \right) = h\left( {2x - 2y} \right)$

$h'\left( {y - x} \right)\left( {y + x} \right) =  - 2h\left( {y - x} \right)$

$h'\underbrace {\left( {y + x} \right)}_t =  - 2h$

$\frac{{dh}}{{dt}}t =  - 2h$

$\frac{{dh}}{h} =  - \frac{{2dt}}{t}{\rm{   }}{\text{         }}{\text{         }}      /\int {} $

$\ln h =  - 2\ln t$

$h = {t^{ - 2}}$

$h = \frac{1}{{{{\left( {x + y} \right)}^2}}}$

 

\[\underbrace {{{\left( {\frac{y}{{x + y}}} \right)}^2}}_{{P_1}}dx + \underbrace {{{\left( {\frac{x}{{x + y}}} \right)}^2}}_{{Q_1}}dy = 0\]

$\frac{{\partial {P_1}}}{{\partial y}} = 2\frac{y}{{x + y}}\frac{{1\left( {x + y} \right) - y \cdot 1}}{{{{\left( {x + y} \right)}^2}}} = 2\frac{y}{{x + y}}\frac{{x + y - y}}{{{{\left( {x + y} \right)}^2}}} = \frac{{2xy}}{{{{\left( {x + y} \right)}^3}}}$

$\frac{{\partial {Q_1}}}{{\partial x}} = 2\frac{x}{{x + y}}\frac{{1\left( {x + y} \right) - x \cdot 1}}{{{{\left( {x + y} \right)}^2}}} = 2\frac{x}{{x + y}}\frac{{x + y - x}}{{{{\left( {x + y} \right)}^2}}} = \frac{{2xy}}{{{{\left( {x + y} \right)}^3}}}$

$\frac{{\partial {P_1}}}{{\partial y}} = \frac{{\partial {Q_1}}}{{\partial x}}$

$\exists F\left( {x,y} \right) = C:{\rm{   }}{\text{         }}{\text{         }}      \underbrace {\frac{{\partial F}}{{\partial x}}}_{{P_1}}dx + \underbrace {\frac{{\partial F}}{{\partial y}}}_{{Q_1}}dy = 0$

 

$\frac{{\partial F}}{{\partial x}} = {P_1} = {\left( {\frac{y}{{x + y}}} \right)^2}{\rm{   }}{\text{         }}{\text{         }}      /\int {} dx$

$F\left( {x,y} \right) = \int {\frac{{{y^2}}}{{{{\left( {x + y} \right)}^2}}}} dx = {y^2}\int {\frac{1}{{{{\left( {x + y} \right)}^2}}}} dx = \left| {\begin{array}{*{20}{c}}
{x + y = t}\\
{dx = dt}
\end{array}} \right| = $

${\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }} {\text{         }} {\text{         }}  ={y^2}\int {{t^{ - 2}}} dt = {y^2}\left( { - \frac{1}{t}} \right) + \varphi \left( y \right)$

$F\left( {x,y} \right) =  - \frac{{{y^2}}}{{x + y}} + \varphi \left( y \right){\rm{   }}{\text{         }} {\text{         }}  /\frac{\partial }{{\partial y}}$

$\frac{{\partial F}}{{\partial y}} = \frac{{ - 2y\left( {x + y} \right) + {y^2} \cdot 1}}{{{{\left( {x + y} \right)}^2}}} + \varphi '\left( y \right) = {Q_1}$

${\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }} {\text{         }}{\text{         }}\frac{{ - 2yx - 2{y^2} + {y^2}}}{{{{\left( {x + y} \right)}^2}}} + \varphi '\left( y \right) = {\left( {\frac{x}{{x + y}}} \right)^2}$

$\varphi '\left( y \right) = {\left( {\frac{x}{{x + y}}} \right)^2} - \frac{{ - 2yx - {y^2}}}{{{{\left( {x + y} \right)}^2}}}$

$\varphi '\left( y \right) = \frac{{{x^2} + 2yx + {y^2}}}{{{{\left( {x + y} \right)}^2}}} = 1$

$\varphi \left( y \right) = y + {C_1}$

\[F\left( {x,y} \right) =  - \frac{{{y^2}}}{{x + y}} + y + {C_1} = \frac{{ - {y^2} + xy + {y^2}}}{{x + y}} + {C_1}\]

\[\frac{{xy}}{{x + y}} = C\]