1.2.2
1/y+1/(3-y)=(3-y)/(y(3-y))+y/(y(3-y))=3/(y(3-y))
なので
y'=2y(1-y/6)
dy/dx=y(1-y/3)
dy/dx=1/3*y(3-y)
3/(y(3-y))*dy=dx
{1/y+1/(3-y)}*dy=dx
両辺を積分すると
∫{1/y+1/(3-y)}dy=∫dx
log|y|-log|3-y|=x+C'
log|y/(3-y)|=x+C'
y/(3-y)=e^(x+C')
y/(3-y)=e^x*e^C'
y/(3-y)=C*e^x
y=C*e^x*(3-y)
y=3C*e^x-y*C*e^x
y+y*C*e^x=3C*e^x
(1+C*e^x)y=3C*e^x
y=3C*e^x/(1+C*e^x)
ただし、C',Cは積分定数でC=e^C'
y(0)=1なので
1=3C*e^0/(1+C*e^0)
1=3C/(1+C)
1+C=3C
2C=1
C=1/2
y=3*1/2*e^x/(1+1/2*e^x)
y=3e^x/(2+e^x)
1.2.4
y>0
dy/dx=(y^2-x^2)/(2xy)
y=xuとおくと
dy/dx=u+x*du/dx
dy/dx=(y^2-x^2)/(2xy)
u+x*du/dx=((xu)^2-x^2)/(2x*xu)
u+x*du/dx=((u^2-1)*x^2)/(2u*x^2)
x*du/dx=(u^2-1)/(2u)-u
x*du/dx=u/2-1/(2u)-u
x*du/dx=-u/2-1/(2u)
x*du/dx=-1/2*(u+1/u)
x*du/dx=-1/2*(u^2+1)/u
2u/(u^2+1)*du=-x*dx
両辺を積分すると
∫{2u/(u^2+1)}du=∫{-x}dx
log|u^2+1|=-1/2*x^2+C'
u^2+1=e^(-1/2*x^2+C')
u^2=-1+e^(-1/2*x^2)*e^C'
u^2=-1+C*e^(-1/2*x^2)
(y/x)^2=-1+C*e^(-1/2*x^2)
y^2=x^2*(-1+C*e^(-1/2*x^2))
ただし、C',Cは積分定数でC=e^C'
y>0なので
y=x*√(-1+C*e^(-1/2*x^2))
(1,1)を通るものは
1=1*√(-1+C*e^(-1/2*1^2))
1=√(-1+C*e^(-1/2))
1=-1+C*e^(-1/2)
C*e^(-1/2)=2
C=2*e^(1/2)
y=x*√(-1+2*e^(1/2)*e^(-1/2*x^2))
y=x*√(-1+2*e^(-1/2*x^2+1/2))
