Линеарне диференцијалне једначине
\[\begin{array}{l} y' + f\left( x \right)y = g\left( x \right) \\ смена:y = u \cdot v{\rm{ }} {\text{ }}{\text{ }}где{\rm{ }}{\text{ }}{\text{ }}су{\rm{ }}{\text{ }}{\text{ }}u = u\left( x \right),v = v\left( x \right) \\ {\rm{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} y' = u'v + uv' \\ u'v + uv' + f\left( x \right)uv = g\left( x \right) \\ u'v + u\left( {v' + f\left( x \right)v} \right) = g\left( x \right) \\ v' + f\left( x \right)v = 0 \wedge u'v = g\left( x \right) \end{array}\]
Задаци
1. Наћи опште решење линеарне диференцијалне једначине $y' - \frac{2}{{x + 1}}y = {\left( {x + 1} \right)^3}$.
Ова линеарна једначина већ је дата у одговарајућем облику где је $f\left( x \right) = - \frac{2}{{x + 1}}$ а $g\left( x \right) = {\left( {x + 1} \right)^3}$.
Па одмах уводимо смену $y = u \cdot v \Rightarrow y' = u'v + uv'$
$u'v + uv' - \frac{2}{{x + 1}}uv = {\left( {x + 1} \right)^3}$
$u'v + u\left( {v' - \frac{2}{{x + 1}}v} \right) = {\left( {x + 1} \right)^3}$
$v' - \frac{2}{{x + 1}}v = 0 {\text{ }}{\text{ }} {\text{ }}\wedge{\text{ }}{\text{ }} {\text{ }} u'v = {\left( {x + 1} \right)^3}$
$v' = \frac{2}{{x + 1}}v$
$\frac{{dv}}{{dx}} = \frac{2}{{x + 1}}v$
$\frac{{dv}}{v} = \frac{2}{{x + 1}}dx{\rm{ }} {\text{ }}{\text{ }} /\int {} $
$\int {\frac{{dv}}{v} = \int {\frac{2}{{x + 1}}dx} } $
${\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} \left| {\begin{array}{*{20}{c}}
{x + 1 = t}\\
{dx = dt}
\end{array}} \right|$
$\ln \left| v \right| = 2\int {\frac{{dt}}{t}} $
$\ln \left| v \right| = 2\ln \left| t \right|$
$v = {\left( {x + 1} \right)^2}$
Када смо израчунали функцију $v$ враћамо се назад у једначину да израчунамо $u$
$u'{\left( {x + 1} \right)^2} = {\left( {x + 1} \right)^3}$
$\frac{{du}}{{dx}} = x + 1$
$du = \left( {x + 1} \right)dx{\rm{ }}{\text{ }}{\text{ }}/\int {} $
$\int {du} = \int {\left( {x + 1} \right)dx} $
$u = \frac{{{x^2}}}{2} + x + C$
\[y = uv = {\left( {x + 1} \right)^2}\left( {\frac{{{x^2}}}{2} + x + C} \right)\]
2. Наћи партикуларно решење диференцијалне једначине $y' + 3{x^2}y = 6{x^2}$ које задовољава почетни услов $y\left( 0 \right) = 3$.
$y = uv \Rightarrow y' = u'v + uv'$
$u'v + uv' + 3{x^2}uv = 6{x^2}$
$u'v + u\left( {v' + 3{x^2}v} \right) = 6{x^2}$
$v' + 3{x^2}v = 0$
$\frac{{dv}}{{dx}} = - 3{x^2}v$
$\frac{{dv}}{v} = - 3{x^2}dx{\rm{ }{\text{ }}{\text{ }} }/\int {} $
$\int {\frac{{dv}}{v}} = \int { - 3{x^2}dx} $
$\ln \left| v \right| = - 3\frac{{{x^3}}}{3}$
$\ln \left| v \right| = - {x^3}$
$v = {e^{ - {x^3}}}$
$u'{e^{ - {x^3}}} = 6{x^2}$
$\frac{{du}}{{dx}} = 6{x^2}{e^{{x^3}}}$
$du = 6{x^2}{e^{{x^3}}}dx{\rm{ }}{\text{ }}{\text{ }} /\int {} $
$\int {du} = \int {6{x^2}{e^{{x^3}}}dx} $
${\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} \left| {\begin{array}{*{20}{c}}
{{x^3} = t}\\
{3{x^2}dx = dt}
\end{array}} \right|$
$u = 2\int {{e^t}dt} $
$u = 2{e^t} + C$
$u = 2{e^{{x^3}}} + C$
\[y = uv = {e^{ - {x^3}}}\left( {2{e^{{x^3}}} + C} \right) = 2 + \frac{C}{{{e^{{x^3}}}}}\]
Уврштавајући почетни услов, односно $x = 0,y = 3$ добијамо
$3 = 2 + \frac{C}{{{e^0}}} \Rightarrow 3 = 2 + \frac{C}{1} \Rightarrow C = 1$
па је партикуларно решење наше диференцијалне једначине
\[y = 2 + \frac{1}{{{e^{{x^3}}}}}\]
3. Наћи опште решење диференцијалне једначине ${x^2}y' + 2xy = {\cos ^2}x$.
Најпре једначину делимо са ${{x^2}}$ да бисмо је свели на облик линеарне диференцијалне једначине.
${x^2}y' + 2xy = {\cos ^2}x{\rm{ }}{\text{ }}{\text{ }} /:{x^2}$
$y' + \frac{2}{x}y = \frac{{{{\cos }^2}x}}{{{x^2}}}$
$y = uv \Rightarrow y' = u'v + uv'$
$u'v + uv' + \frac{2}{x}uv = \frac{{{{\cos }^2}x}}{{{x^2}}}$
$u'v + u\left( {v' + \frac{2}{x}v} \right) = \frac{{{{\cos }^2}x}}{{{x^2}}}$
$v' + \frac{2}{x}v = 0$
$\frac{{dv}}{{dx}} = - \frac{2}{x}v$
$\frac{{dv}}{v} = - \frac{2}{x}dx{\rm{ }}{\text{ }}{\text{ }} /\int {} $
$\int {\frac{{dv}}{v}} = \int { - \frac{2}{x}dx} $
$\ln \left| v \right| = - 2\ln \left| x \right|$
$v = \frac{1}{{{x^2}}}$
$u'\frac{1}{{{x^2}}} = \frac{{{{\cos }^2}x}}{{{x^2}}}$
$\frac{{du}}{{dx}} = {\cos ^2}x$
$du = {\cos ^2}xdx{\rm{ }} {\text{ }}{\text{ }} /\int {} $
Како је $\cos \frac{x}{2} = \sqrt {\frac{{1 + \cos x}}{2}} \Rightarrow \cos x = \sqrt {\frac{{1 + \cos 2x}}{2}} \Rightarrow {\cos ^2}x = \frac{{1 + \cos 2x}}{2}$
$u = \int {{{\cos }^2}xdx} = \int {\frac{{1 + \cos 2x}}{2}} dx = \int {\frac{1}{2}} dx + \int {\frac{1}{2}} \cos 2xdx = $
${\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} \left| {\begin{array}{*{20}{c}}
{2x = t}\\
{2dx = dt}\\
{dx = \frac{{dt}}{2}}
\end{array}} \right|$
$ {\text{ }} {\text{ }} {\text{ }} = \frac{1}{2}x + \frac{1}{2}\int {\cos t} \frac{{dt}}{2} = \frac{1}{2}x + \frac{1}{4}\sin 2x + C$
\[y = uv = \frac{1}{{{x^2}}}\left( {\frac{1}{2}x + \frac{1}{4}\sin 2x + C} \right) = \frac{1}{{2x}} + \frac{1}{{4{x^2}}}\sin 2x + \frac{C}{{{x^2}}}\]
4. Наћи опште решење диференцијалне једначине $y' + y\cos x = \frac{1}{2}\sin 2x$.
$y = uv \Rightarrow y' = u'v + uv'$
$u'v + uv' + uv\cos x = \frac{1}{2}\sin 2x$
$u'v + u\left( {v' + v\cos x} \right) = \frac{1}{2}\sin 2x$
$v' + v\cos x = 0$
$\frac{{dv}}{{dx}} = - v\cos x$
$\frac{{dv}}{v} = - \cos xdx{\rm{ }}{\text{ }}{\text{ }} /\int {} $
$\ln \left| v \right| = - \int {\cos xdx} $
$\ln \left| v \right| = - \sin x$
$v = {e^{ - \sin x}}$
$u'{e^{ - \sin x}} = \frac{1}{2}2\sin x\cos x$
$\frac{{du}}{{dx}}{e^{ - \sin x}} = \sin x\cos x$
$du = \sin x\cos x{e^{\sin x}}dx{\rm{ }}{\text{ }}{\text{ }} /\int {} $
$u = \int {\sin x\cos x{e^{\sin x}}dx} $
${\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} \left| {\begin{array}{*{20}{c}}
{\sin x = t}\\
{\cos xdx = dt}
\end{array}} \right|$
$u = \int {t{e^t}dt} = $
${\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} \left| {\begin{array}{*{20}{c}}
{u = t}&{dv = {e^t}dt}\\
{du = dt}&{v = {e^t}}
\end{array}} \right|$
$ {\text{ }} {\text{ }} {\text{ }} = t{e^t} + \int {{e^t}dt} = t{e^t} + {e^t} + C$
$ {\text{ }} {\text{ }} {\text{ }} = \sin x{e^{\sin x}} + {e^{\sin x}} + C$
\[y = uv = {e^{ - \sin x}}\left( {\sin x{e^{\sin x}} + {e^{\sin x}} + C} \right) = \sin x + 1 + \frac{C}{{{e^{\sin x}}}}\]
5. Наћи оно партикуларно решење диференцијалне једначине $dy = \frac{{{x^3} + y}}{x}dx$ које задовољава почетни услов $y\left( 2 \right) = 8$.
$dy = \frac{{{x^3} + y}}{x}dx{\rm{ }}{\text{ }} {\text{ }} /:dx$
$\frac{{dy}}{{dx}} = {x^2} + \frac{y}{x}$
$y' = {x^2} + \frac{1}{x}y$
$y' - \frac{1}{x}y = {x^2}$
$y = uv \Rightarrow y' = u'v + uv'$
$u'v + uv' - \frac{1}{x}uv = {x^2}$
$u'v + u\left( {v' - \frac{1}{x}v} \right) = {x^2}$
$v' - \frac{1}{x}v = 0$
$\frac{{dv}}{{dx}} = \frac{1}{x}v$
$\frac{{dv}}{v} = \frac{1}{x}dx{\rm{ }}{\text{ }} {\text{ }} /\int {} $
$\ln \left| v \right| = \ln \left| x \right|$
$v = x$
$u'x = {x^2}$
$\frac{{du}}{{dx}} = x$
$du = xdx{\rm{ }} {\text{ }} {\text{ }} /\int {} $
$u = \frac{{{x^2}}}{2} + C$
\[y = uv = \frac{{{x^3}}}{2} + Cx\]
\[8 = \frac{{{2^3}}}{2} + 2C \Rightarrow C = 2\]
\[y = \frac{{{x^3}}}{2} + 2x\]
6. Наћи опште решење диференцијалне једначине $\sin xy' + \cos xy = {\sin ^2}x$.
$y' + \frac{{\cos x}}{{\sin x}}y = \sin x$
$y' + ctgxy = \sin x$
$y = uv \Rightarrow y' = u'v + uv'$
$u'v + uv' + ctgxuv = \sin x$
$u'v + u\left( {v' + ctgxv} \right) = \sin x$
$v' + ctgxv = 0$
$\frac{{dv}}{{dx}} = - ctgxv$
$\frac{{dv}}{v} = - \frac{{\cos x}}{{\sin x}}dx{\rm{ }}{\text{ }} {\text{ }} /\int {} $
$\ln \left| v \right| = - \int {\frac{{\cos x}}{{\sin x}}dx} $
${\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }}{\text{ }} \left| {\begin{array}{*{20}{c}}
{\sin x = t}\\
{\cos xdx = dt}
\end{array}} \right|$
$\ln \left| v \right| = - \int {\frac{{dt}}{t}} $
$\ln \left| v \right| = - \ln \left| t \right|$
$v = \frac{1}{t}$
$v = \frac{1}{{\sin x}}$
$u'\frac{1}{{\sin x}} = \sin x$
$\frac{{du}}{{dx}} = {\sin ^2}x$
$du = {\sin ^2}xdx{\rm{ }}{\text{ }} {\text{ }} /\int {} $
$u = \int {{{\sin }^2}xdx} = {\int {\sqrt {\frac{{1 - \cos 2x}}{2}} } ^2}dx = \int {\frac{1}{2}} dx - \int {\frac{{\cos 2x}}{2}} dx = $
${\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} \left| {\begin{array}{*{20}{c}}
{2x = t}\\
{2dx = dt}\\
{dx = \frac{{dt}}{2}}
\end{array}} \right|$
$ {\text{ }} {\text{ }} {\text{ }} = \frac{1}{2}x - \frac{1}{4}\sin 2x + C$
\[y = uv = \frac{1}{{\sin x}}\left( {\frac{1}{2}x - \frac{1}{4}\sin 2x + C} \right) = \frac{x}{{2\sin x}} - \frac{{\cos x}}{2} + \frac{C}{{\sin x}}\]
7. Наћи опште решење диференцијалне једначине $\left( {1 - {y^2}} \right)dx + \left( {1 - xy} \right)dy = 0$.
Ако бисмо ову диференцијалну једначину покушали свести на линеарну диференцијалну једначину облика $y' + f\left( x \right)y = g\left( x \right)$, наишли бисмо на проблем јер не бисмо могли раздвојити $x$ и $y$.
Из тог разлога, прелазимо на облик $x' + f\left( y \right)x = g\left( y \right)$, односно на линеарну диференцијалну једначину по $x$.
$\left( {1 - {y^2}} \right)dx = - \left( {1 - xy} \right)dy$
$\frac{{dx}}{{dy}} = \frac{{xy - 1}}{{1 - {y^2}}}$
$x' = \frac{y}{{1 - {y^2}}}x - \frac{1}{{1 - {y^2}}}$
$x' - \frac{y}{{1 - {y^2}}}x = \frac{1}{{{y^2} - 1}}$
Уводимо смену $x = uv \Rightarrow x' = u'v + uv'$ где су $u = u\left( y \right)$ и $v = v\left( y \right)$.
$u'v + uv' - \frac{y}{{1 - {y^2}}}uv = \frac{1}{{{y^2} - 1}}$
$u'v + u\left( {v' - \frac{y}{{1 - {y^2}}}v} \right) = \frac{1}{{{y^2} - 1}}$
$v' - \frac{y}{{1 - {y^2}}}v = 0$
$\frac{{dv}}{{dy}} = \frac{y}{{1 - {y^2}}}v$
$\frac{{dv}}{v} = \frac{y}{{1 - {y^2}}}dy{\rm{ }}{\text{ }} {\text{ }} /\int {} $
$\ln \left| v \right| = \int {\frac{y}{{1 - {y^2}}}dx} $
${\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} \left| {\begin{array}{*{20}{c}}
{1 - {y^2} = t}\\
{ - 2ydy = dt}\\
{ydy = - \frac{{dt}}{2}}
\end{array}} \right|$
$\ln \left| v \right| = \int {\frac{{ - \frac{{dt}}{2}}}{t}} = - \frac{1}{2}\ln \left| t \right| = - \frac{1}{2}\ln \left| {1 - {y^2}} \right|$
$v = \frac{1}{{\sqrt {1 - {y^2}} }}$
$u'\frac{1}{{\sqrt {1 - {y^2}} }} = \frac{1}{{{y^2} - 1}}{\rm{ }}{\text{ }} {\text{ }} / \cdot \sqrt {1 - {y^2}} $
$ u'= - \frac{{\sqrt {1 - {y^2}} }}{{1 - {y^2}}}$
$ \frac{{du}}{{dy}} = - \frac{1}{{\sqrt {1 - {y^2}} }}$
$du = - \frac{1}{{\sqrt {1 - {y^2}} }}dy{\rm{ }} {\text{ }} {\text{ }} /\int {} $
$ u= - arcsiny + C$
\[x = uv = \frac{{ - arcsiny + C}}{{\sqrt {1 - {y^2}} }}\]
8. Наћи оно партикуларно решење диференцијалне једначине ${x^2}y' + xy = 1$ које задовољава почетни услов $y\left( 1 \right) = 2$.
${x^2}y' + xy = 1{\rm{ }}{\text{ }} {\text{ }} /:{x^2}$
$y' + \frac{1}{x}y = \frac{1}{{{x^2}}}$
$y = uv \Rightarrow y' = u'v + uv'$
$u'v + uv' + \frac{1}{x}uv = \frac{1}{{{x^2}}}$
$u'v + u\left( {v' + \frac{1}{x}v} \right) = \frac{1}{{{x^2}}}$
$v' + \frac{1}{x}v = 0$
$\frac{{dv}}{{dx}} = - \frac{1}{x}v$
$\frac{{dv}}{v} = - - \frac{1}{x}dx{\rm{ }}{\text{ }} {\text{ }} /\int {} $
$\ln \left| v \right| = - \ln \left| x \right|$
$v = \frac{1}{x}$
$u'\frac{1}{x} = \frac{1}{{{x^2}}}$
$\frac{{du}}{{dx}} = \frac{1}{x}$
$du = \frac{1}{x}dx{\rm{ }} {\text{ }} {\text{ }} /\int {} $
$u = \ln \left| x \right| + \ln C$
$u = \ln xC$
\[y = uv = \frac{{\ln xC}}{x}\]
\[2 = \frac{{\ln 1 \cdot C}}{1} \Rightarrow \ln C = 2 \Rightarrow C = {e^2}\]
\[y = \frac{{\ln x + 2}}{x}\]
9. Наћи оно партикуларно решење диференцијалне једначине $xy' - \frac{y}{{x + 1}} = x$ које пролази кроз тачку $T\left( {1, - 1} \right)$.
$xy' - \frac{y}{{x + 1}} = x{\rm{ }}{\text{ }} {\text{ }} /:x$
$y' - \frac{y}{{x\left( {x + 1} \right)}} = 1$
$y = uv \Rightarrow y' = u'v + uv'$
$u'v + uv' - \frac{1}{{x\left( {x + 1} \right)}}uv = 1$
$u'v + u\left( {v' - \frac{1}{{x\left( {x + 1} \right)}}v} \right) = 1$
$v' - \frac{1}{{x\left( {x + 1} \right)}}v = 0$
$\frac{{dv}}{{dx}} = \frac{1}{{x\left( {x + 1} \right)}}v$
$\frac{{dv}}{v} = \frac{1}{{x\left( {x + 1} \right)}}dx{\rm{ }}{\text{ }} {\text{ }} /\int {} $
$\ln \left| v \right| = \int {\frac{1}{{{x^2} + x}}dx} = \int {\frac{1}{{{x^2} + x + \frac{1}{4} - \frac{1}{4}}}dx} = \int {\frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}}}} dx = $
${\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }}{\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} \left| {\begin{array}{*{20}{c}}
{x + \frac{1}{2} = t}\\
{dx = dt}
\end{array}} \right|$
$ {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} = \int {\frac{1}{{{t^2} - {{\left( {\frac{1}{2}} \right)}^2}}}} dt = \frac{1}{{2 \cdot \frac{1}{2}}}\ln \left| {\frac{{t - \frac{1}{2}}}{{t + \frac{1}{2}}}} \right| = \ln \left| {\frac{x}{{x + 1}}} \right|$
$v = \frac{x}{{x + 1}}$
$u'\frac{x}{{x + 1}} = 1$
$\frac{{du}}{{dx}} = \frac{{x + 1}}{x}$
$du = \left( {1 + \frac{1}{x}} \right)dx{\rm{ }} {\text{ }} {\text{ }} /\int {} $
$u = x + \ln x + C$
\[y = uv = \frac{x}{{x + 1}}\left( {x + \ln x + C} \right)\]
\[ - 1 = \frac{1}{{1 + 1}}\left( {1 + \ln 1 + C} \right) \Rightarrow - 1 = \frac{1}{2} + \frac{C}{2} \Rightarrow C = - 3\]
\[y = \frac{x}{{x + 1}}\left( {x + \ln x - 3} \right)\]
10. Наћи оно партикуларно решење диференцијалне једначине $\left( {1 - {x^2}} \right)y' + xy - 1 = 0$ које пролази кроз тачку $T\left( {0, 1} \right)$.
$\left( {1 - {x^2}} \right)y' + xy - 1 = 0{\rm{ }{\text{ }} {\text{ }} }/:\left( {1 - {x^2}} \right)$
$y' + \frac{x}{{1 - {x^2}}}y = \frac{1}{{1 - {x^2}}}$
$y = uv \Rightarrow y' = u'v + uv'$
$u'v + uv' + \frac{x}{{1 - {x^2}}}uv = \frac{1}{{1 - {x^2}}}$
$u'v + u\left( {v' + \frac{x}{{1 - {x^2}}}v} \right) = \frac{1}{{1 - {x^2}}}$
$v' + \frac{x}{{1 - {x^2}}}v = 0$
$\frac{{dv}}{{dx}} = - \frac{x}{{1 - {x^2}}}v$
$\frac{{dv}}{v} = - \frac{x}{{1 - {x^2}}}dx{\rm{ }}{\text{ }} {\text{ }} /\int {} $
$\ln \left| v \right| = \int { - \frac{x}{{1 - {x^2}}}dx} = $
$ {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} \left| {\begin{array}{*{20}{c}}
{1 - {x^2} = t}\\
\begin{array}{l}
- 2xdx = dt\\
- xdx = \frac{{dt}}{2}
\end{array}
\end{array}} \right|$
$ {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} {\text{ }} = \int {\frac{{\frac{{dt}}{2}}}{t}} = \frac{1}{2}\ln \left| t \right| = \ln \sqrt {1 - {x^2}} $
$v = \sqrt {1 - {x^2}} $
$u'\sqrt {1 - {x^2}} = \frac{1}{{1 - {x^2}}}$
$\frac{{du}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }}\frac{1}{{1 - {x^2}}}$
$du = \frac{1}{{{{\sqrt {1 - {x^2}} }^3}}}dx{\rm{ }}{\text{ }} {\text{ }} /\int {} $
$u = \int {\frac{1}{{{{\sqrt {1 - {x^2}} }^3}}}dx} = \left| {\begin{array}{*{20}{c}}
{x = \sin t}\\
{dx = \cos tdt}
\end{array}} \right| = \int {\frac{{\cos tdt}}{{{{\sqrt {1 - {{\sin }^2}t} }^3}}}} = \int {\frac{{\cos tdt}}{{{{\sqrt {{{\cos }^2}t} }^3}}}} = \int {\frac{{dt}}{{{{\cos }^2}t}}} =$
${\text{ }} {\text{ }} = tgt + C = tg\left( {\arcsin x} \right) + C = \frac{x}{{\sqrt {1 - {x^2}} }} + C$
\[y = uv = x + C\sqrt {1 - {x^2}} \]
\[1 = 0 + C\sqrt {1 - {0^2}} \Rightarrow C = 1\]
\[y = x + \sqrt {1 - {x^2}} \]
11. Наћи оно партикуларно решење диференцијалне једначине $y' = \frac{y}{{2y\ln y + y - x}}$ које пролази кроз тачку $T\left( {1, 1} \right)$.
$\frac{{dy}}{{dx}} = \frac{y}{{2y\ln y + y - x}}$
$\frac{{dx}}{{dy}} = \frac{{2y\ln y + y - x}}{y}$
$x' = 2\ln y + 1 - \frac{x}{y}$
$x' + \frac{1}{y}x = 2\ln y + 1$
$x = uv \Rightarrow x' = u'v + uv'$
$u'v + uv' + \frac{1}{y}uv = 2\ln y + 1$
$u'v + u\left( {v' + \frac{1}{y}v} \right) = 2\ln y + 1$
$v' + \frac{1}{y}v = 0$
$\frac{{dv}}{{dy}} = - \frac{1}{y}v$
$\frac{{dv}}{v} = - \frac{1}{y}dy{\rm{ }}{\text{ }} {\text{ }} /\int {} $
$\ln \left| v \right| = - \ln \left| y \right|$
$v = \frac{1}{y}$
$u'\frac{1}{y} = 2\ln y + 1$
$\frac{{du}}{{dy}} = 2y\ln y + y$
$du = \left( {2y\ln y + y} \right)dy{\rm{ }}{\text{ }} {\text{ }} /\int {} $
$u = 2\int {y\ln ydy + \int {ydy} } = $
${\text{ }} {\text{ }} \left| {\begin{array}{*{20}{c}}
{u = \ln y}&{dv = ydy}\\
{du = \frac{1}{y}dy}&{v = \frac{{{y^2}}}{2}}
\end{array}} \right|$
$ {\text{ }} {\text{ }} = 2\left( {\frac{{{y^2}}}{2}\ln y - \int {\frac{{{y^2}}}{2}\frac{1}{y}dy} } \right) + \frac{{{y^2}}}{2} = $
$ {\text{ }} {\text{ }} = {y^2}\ln y - \frac{{{y^2}}}{2} + \frac{{{y^2}}}{2} + C = {y^2}\ln y + C$
\[x = uv = \frac{1}{y}\left( {{y^2}\ln y + C} \right) = y\ln y + \frac{C}{y}\]
\[1 = 1\ln 1 + \frac{C}{1} \Rightarrow C = 1\]
\[x = y\ln y + \frac{1}{y}\]