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(1)
dy/dx = 10^(x+y)
∫10^(-y) dy =∫10^x dx
-(1/ln10) 10^(-y) = (1/ln10).10^x +C'
10^(-y) = -10^x -C'.ln10
-y = C''.10^{-10^x]
y = C.10^{-10^x]
(2)
dy/dx +y = e^(-x)
e^x.[dy/dx +y] = 1
d/dx ( e^x. y) =1
e^x. y = x +C
y = (x +C) .e^(-x)
(3)
dy/dx +2xy =4x
e^(x^2).[dy/dx +2xy] =4x.e^(x^2)
d/dx [ e^(x^2).y] =4x.e^(x^2)
e^(x^2).y =∫4x.e^(x^2) dx
=2e^(x^2) + C
y =2 + C.e^(-x^2)
dy/dx = 10^(x+y)
∫10^(-y) dy =∫10^x dx
-(1/ln10) 10^(-y) = (1/ln10).10^x +C'
10^(-y) = -10^x -C'.ln10
-y = C''.10^{-10^x]
y = C.10^{-10^x]
(2)
dy/dx +y = e^(-x)
e^x.[dy/dx +y] = 1
d/dx ( e^x. y) =1
e^x. y = x +C
y = (x +C) .e^(-x)
(3)
dy/dx +2xy =4x
e^(x^2).[dy/dx +2xy] =4x.e^(x^2)
d/dx [ e^(x^2).y] =4x.e^(x^2)
e^(x^2).y =∫4x.e^(x^2) dx
=2e^(x^2) + C
y =2 + C.e^(-x^2)
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谢谢大侠!辛苦了!只是第一题是dx/dy 该怎么解?
大侠,e^x.[dy/dx +y] = 1 这步怎么来的看不懂啊
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