2013/11-2013/1 Yuji.W |
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◎ 1階 x,y
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■ y;=f(a*x+b*y+c) a*x+b*y+c=u(x) と置くと、 a+b*y;=u; y;=(u;-a)/b (u;-a)/b=f(u) u;=b*f(u)+a ★. [1/(b*f+a)]}*du=1*dx 変数分離 ● ${[1/(x^2+a^2)]*dx}=(1/a)*arctan(x/a)+C ★ y;=(x+y+1)^2 y(0)=0 x+y+1=u と置くと 1+y;=u; u(0)=1 u;=u^2+1 [1/(u^2+1)]*du=dx arctan(u)=x+Pi/4 u=tan(x+Pi/4) y=tan(x+Pi/4)-x-1 ★ (x-y)*y;=1 置く[u=x-y-1] y;=1-u; (u+1)*(1-u;)=1 (u+1)*u;=u+1-1=u (1+1/u)*u;=1 u+Lu=x+C (x-y-1)+ln(x-y-1)=x+C ln(x-y-1)=C+y x-y-1=C*Ey |
☆ 2013 Yuji.W ☆