求救 工程數學微分方程兩個問題(急)

1.dy/dx=(y^2-y-2)/(x^2-2x+1), y(4)=4dy/(y^2-y-2)=dx/(x^2-2x+1)dy/(y+1)(y-2)=dx/(x-1)^2
∫dy/3(y-2)-∫dy/3(y+1)=∫d(x-1)/(x-1)^2[ln(y-2)-ln(y+1)]/3=-1/(x-1)+c{ln[(y-2)/(y+1)]}=-3/(x-1)+3cx=y=4 => 3c=[ln(2/5)]+1=0.0837=> (y-2)/(y+1)=e^[-3/(x-1)+0.0837]......ans
2.y"+4y=1/cos2x, y(0)=0, y'(0)=00=D^2+4 => D=+-2jyh(x)=a*cos(2x)+b*sin(2x)參變數法: y1=cos(2x), y2=sin(2x)[y1 .y2.]{A'}.{..0...}
[y1' y2']{B'}={sec(x)}w=|y1 .y2.|
..[y1' y2'|=y1*y2'-y2*y1'=2[cos^2(2x)+sin^2(2x)]=2
A'=-y2*f(x)/wA=-∫sin(2x)dx/2cos(2x)=-∫sin(2x)d(2x)/4cos(2x)=∫d(cos(2x))/4cos(2x)=ln(cos(2x))/4
B'= y1*f(x)/wB=∫cos(2x)dx/2cos(2x)=∫dx/2=x/2=> yp(x)=A*y1+B*y2=cos(2x)ln(cos(2x))/4+xsin(2x)/2=> y(x)=yh(x)+yp(x)=[a*cos(2x)+b*sin(2x)]+[cos(2x)ln(cos(2x))/4+xsin(2x)/2]......ans

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