Matematika/Pirmos eilės tiesinės diferencialinės lygtys

Šis straipsnis yra apie Pirmos eilės tiesines diferencialines lygtis.

• $y'+P(x)y=Q(x),$

${dy \over dx}+P(x)y=Q(x),$

$y=uv$, $y'=u'v+uv'.$

$u'v+uv'+P(x)uv=Q(x),$

$v(u'+P(x)u)+uv'=Q(x);$

$u'+P(x)u=0,$

${du \over u}=-P(x)dx,$

$u=C_{1}e^{-\int P(x)dx};$

$C_{1}e^{-\int P(x)dx}v'=Q(x),$

$v'={1 \over C_{1}}Q(x)e^{\int P(x)dx},$

$v=\int {1 \over C_{1}}Q(x)e^{\int P(x)dx}dx+C_{2};$

$y=uv=C_{1}e^{-\int P(x)dx}(\int {1 \over C_{1}}Q(x)e^{\int P(x)dx}dx+C_{2})=e^{-\int P(x)dx}(\int Q(x)e^{\int P(x)dx}dx+C_{1}C_{2})=$

$=e^{-\int P(x)dx}(\int Q(x)e^{\int P(x)dx}dx+C).$

• $y'={2y \over x}+x^{2}e^{x}-1,$

$y'-{2 \over x}\cdot y=x^{2}e^{x}-1,$

$y=uv,$ $y'=u'v+uv',$

$u'v+uv'-{2 \over x}uv=x^{2}e^{x}-1,$

$v(u'-{2 \over x}u)+uv'=x^{2}e^{x}-1;$

$u'-{2 \over x}u=0,$

${du \over u}={2 \over x}dx,$

$\ln |u|=2\ln |x|,$

$u=x^{2};$

$x^{2}v'=x^{2}e^{x}-1,$

$dv=(e^{x}-{1 \over x^{2}})dx,$

$v=e^{x}+{1 \over x}+C;$

$y=uv=x^{2}(e^{x}+{1 \over x}+C)=Cx^{2}+x^{2}e^{x}+x.$

• $y'-ay=e^{bx},$

$y=uv,$ $y'=u'v+uv',$

$u'v+uv'-auv=e^{bx},$

$v(u'-au)+uv'=e^{bx};$

$u'-au=0,$

${du \over dx}=au,$

${du \over u}=a\;dx,$

$\ln |u|=ax,$

$u=e^{ax};$

$e^{ax}v'=e^{bx},$

${dv \over dx}=e^{bx-ax},$

$\int dv=\int e^{(b-a)x}dx,$

$v={1 \over b-a}e^{(b-a)x}+C,$ jei $a\neq b$ ir $v=x+C,$ jei $a=b,$ nes $e^{0}=1;$

$y=uv=e^{ax}({e^{(b-a)x} \over b-a}+C)={e^{bx} \over b-a}+Ce^{ax},$ jei $a\neq b$ ir $y=e^{ax}(x+C),$ jei $a=b.$

• $x_{y}'+P(y)x=Q(y),$

${dx \over dy}+P(y)x=Q(y),$

$x=uv,$ $u=u(y),$ $v=v(y).$

• $y'={1 \over \cos ^{2}y-x\tan y},$

${1 \over y_{x}'}=x_{y}',$

${1 \over x_{y}'}={1 \over \cos ^{2}y-x\tan y},$

$x_{y}'=\cos ^{2}y-x\tan y,$

$x_{y}'+x\tan y=\cos ^{2}y,$

$x=x(y),$ $x=uv,$ $x_{y}'=u'v+uv'=u'(x)v(x)+u(x)v'(x),$

$u'v+uv'+uv\tan y=\cos ^{2}y,$

$v(u'+u\tan y)+uv'=\cos ^{2}y;$

$u'+u\tan y=0,$

${du \over u}=-\tan ydy,$

$\ln |u|=\ln |\cos y|;$

$v'\cos y=\cos ^{2}y,$

$v'=\cos y,$

$v=\sin y+C;$

$x=uv=\cos y(\sin y+C)=\sin y\cos y+C\cos y.$

Konstantos variacijos metodas (Lagranžo metodas)

• $y'+P(x)y=Q(x);$

$y'+P(x)y=0,$

${\frac {y'}{y}}=-P(x),$

$\ln y+\ln C=-\int P(x)dx,$

$y=Ce^{-\int P(x)dx};$

$y=C(x)e^{-\int P(x)dx};$

$y'+P(x)y=C'(x)e^{-\int P(x)dx}+C(x)e^{-\int P(x)dx}\cdot (-P(x))+P(x)C(x)e^{-\int P(x)dx}=Q(x),$

$C'(x)e^{-\int P(x)dx}=Q(x),$

$C'(x)=Q(x)e^{\int P(x)dx},$

$C(x)=\int Q(x)e^{\int P(x)dx}dx+C.$

• $y'-{2xy \over 1+x^{2}}=1+x^{2},$ $y|_{x=2}=5;$

$y'-{2xy \over 1+x^{2}}=0,$

${dy \over dx}={2xy \over 1+x^{2}},$

$\int {dy \over y}=\int {2x \over 1+x^{2}}dx,$

$\int {dy \over y}=\int {d(1+x^{2}) \over 1+x^{2}},$

$\ln |y|=\ln |1+x^{2}|+\ln |C|,$

$y=C(1+x^{2});$

$y=C(x)(1+x^{2}),$ $y'=C'(x)\cdot (1+x^{2})+C(x)2x;$

$y'-{2xy \over 1+x^{2}}=C'(x)(1+x^{2})+C(x)\cdot 2x-{2xC(x)(1+x^{2}) \over 1+x^{2}}=1+x^{2},$

$C'(x)\cdot (1+x^{2})=1+x^{2},$

$C'(x)=1,\;C(x)=x+C;$

$y=(x+C)(1+x^{2});$

$5=(2+C)(1+2^{2}),$

$1=2+C,$

$C=-1;$

$y=(x-1)(1+x^{2}).$

• ${dz \over dx}+{3 \over x}z=0,$

$\int {dz \over z}=-3\int {dx \over x},$

$\ln |z|=-3\ln |x|+\ln |C|=\ln |Cx^{-3}|,$

$z={C \over x^{3}};$

$C=C(x),$ $z=C(x)x^{-3},$ ${dz \over dx}={dC(x) \over dx}{1 \over x^{3}}-{3C(x) \over x^{4}},$

${dz \over dx}+{3 \over x}z={dC(x) \over dx}{1 \over x^{3}}-{3C(x) \over x^{4}}+{3 \over x}{C(x) \over x^{3}}=x^{3},$

${dC(x) \over dx}{1 \over x^{3}}=x^{3},$

${dC(x) \over dx}=x^{6},$

$\int dC(x)=\int x^{6}dx,$

$C(x)={x^{7} \over 7}+C_{1};$

$z={C(x) \over x^{3}}={{x^{7} \over 7}+C_{1} \over x^{3}}={x^{4} \over 7}+{C_{1} \over x^{3}}.$