推 EZ0928 :謝謝F大 >< 12/05 22:25
※ 引述《EZ0928 (傲熊)》之銘言:
: y''+(y')^2+1=0
: solve the given differential equation by using u=y'
u'=y''
u'+u^2+1=0
u'= - (u^2+1)
du/dx = - (u^2+1)
du/(1+u^2) = - dx
-1
tan u = - x + C1
u = tan( - x + C1 )
dy/dx = tan( - x + C1 ) = - tan( x - C1 )
dy = - tan( x - C1 ) dx = - tan( x - C1 ) d( x - C1 )
y(x) = ㏑| cos( x - C1 ) | + C2
y(C1) = ㏑| cos( C1 - C1 ) | + C2 = C2
y'(x) = tan( - x + C1 )
y'(C1) = tan( - C1 + C1 ) = tan0 = 0
Ans: y(x) = ㏑| cos( x - C1 ) | + y(C1) ; y(C1) = C2 ; y'(C1) = 0
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※ 編輯: Frobenius 來自: 218.166.100.20 (12/05 02:00)