Matematika/Homogeninės diferencialinės lygtys

Šis straipsnis yra apie Homogeninių diferencialinių lygtčių sprendimą.

• $x^{2}dy-y(y+x)dx=0,$

${dy \over dx}={y(y+x) \over x^{2}},\;x\neq 0,$

$u'x+u={ux(ux+x) \over x^{2}},$ $y=ux,$ $y'=u'x+u,$

$u'x+u=u^{2}+u,$

$u'x=u^{2},$

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

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

$-{1 \over u}=\ln |x|+\ln C,$

$e^{-{x \over y}}=Cx.$

• $y'x-x^{2}-y=0,$

$y=ux,$ $y'=u'x+u,$

$(u'x+u)x-x^{2}-ux=0,$

$u'x+u-x-u=0,$

$u'x=x,$

${du \over dx}=1,$

$\int du=\int dx,$

$u=x+C,$

${y \over x}=x+C,$

$y=x^{2}+Cx.$

• $xy'=y\ln {y \over x},\;y(1)={\sqrt {e}},$

$y'={y \over x}\ln {y \over x},$

$y=ux,$ $y'=u'x+u,$

$u'x+u=u\ln u,$

$u'x=u(\ln u-1),$

${du \over u(\ln u-1)}={dx \over x},$

$\int {d(\ln u-1) \over \ln u-1}=\int {dx \over x},$

$\ln |\ln u-1|=\ln |x|+\ln C,$

$\ln u-1=Cx,$

$u=e^{Cx+1},$

$y=xe^{Cx+1},$

$e^{1 \over 2}=1\cdot e^{C\cdot 1+1},$

$C=-0.5,$

$y=xe^{1-{x \over 2}}.$

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

${y \over x}=u,$ $y=ux,$ ${dy \over dx}=u+x{du \over dx},$

$u+x{du \over dx}={ux^{2} \over {y^{2} \over u^{2}}-u^{2}x^{2}},$

$u+x{du \over dx}={u \over {y^{2} \over x^{2}u^{2}}-u^{2}},$

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

$u+x{du \over dx}={u \over 1-u^{2}},$

$x{du \over dx}={u^{3} \over 1-u^{2}},$

${1-u^{2} \over u^{3}}du={dx \over x},$

$\int ({1 \over u^{3}}-{1 \over u})du=\int {dx \over x},$

$-{1 \over 2u^{2}}-\ln |u|=\ln |x|+\ln |C|,$

$-{1 \over 2u^{2}}=\ln |uxC|,$

$-{x^{2} \over 2y^{2}}=\ln |Cy|,$

$x=y{\sqrt {-2\ln |Cy|}}.$