dy/dx=e^(x-y-2),y(0)=0的特解
dy/dx=e^(x-y-2)
dy/dx=e^-2*e^x/e^-y
e^ydy=e^-2e^xdx
兩邊積分得到
e^y=e^-2e^x+C
代入(0,0)
1=e^-2+C=0
e^y=e^-2e^x+1-e^-2
已知y^2(1+x^2)=c求dy/dx
dy²;(1+x²;)=dc
y²;d(1+x²;)+(1+x²;)dy²;=0
2xy²;dx+2y(1+x²;)dy=0
所以dy/dx=-xy/(1+x²;)
y=x^2(√x-sin2x),求dy
dy=[2x(√x-sin2x)+x^2(1/(2√x)-2cos2x]dx
只要對y求下導再加個dx就行了.
y=sin2x,求dy
dy=cos2xd2x
=2cos2xdx