Matematika/Bernulio diferencialinė lygtis

Bernulio diferencialinė lygtis

• • $y'+P(x)y=Q(x)y^{m},$

$y^{-m}y'+P(x)y^{1-m}=Q(x),$

$y^{1-m}=z,$ $z'=(1-m)y^{-m}y',$

$(1-m)y^{-m}y'+(1-m)P(x)y^{1-m}=(1-m)Q(x),$

$z'+(1-m)P(x)z=(1-m)Q(x).$

Bernulio lygtį galima spręsti panašiai kaip ir pirmosios eilės tiesinę, naudojant keitinį $y=uv.$

• $xy'+y+x^{2}y^{2}e^{x}=0,$

$y'+{y \over x}=-xy^{2}e^{x};$

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

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

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

$u'+{u \over x}=0,$

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

$\int {du \over u}=-\int {dx \over x},$

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

$u={1 \over x};$

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

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

$\int {dv \over v^{2}}=-\int e^{x}dx,$

$-{1 \over v}=-(e^{x}+C),$

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

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

• ${dy \over dx}-{3 \over x}y=-x^{3}y^{2},$

${1 \over y^{2}}{dy \over dx}-{3 \over x}{1 \over y}=-x^{3},$

${1 \over y}=z,\;-{1 \over y^{2}}{dy \over dx}={dz \over dx},$

${dz \over dx}+{3 \over x}z=x^{3};$

šią tiesinę lygtį integruosime variacijos metodu,

${dz \over dx}+{3 \over x}z=0,$

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

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

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

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

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

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

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

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

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

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

$z={1 \over y},\;{1 \over y}={x^{4} \over 7}+{C_{1} \over x^{3}},$

$y={1 \over {x^{4} \over 7}+{C_{1} \over x^{3}}}={1 \over {x^{7}+7C_{1} \over 7x^{3}}}={7x^{3} \over x^{7}+7C_{1}}.$