Given dy/dx in terms of x and y, assume y^2ln(x+1)-y/(e^x+sinx)+2cosx.ln(x^2+1)=1 and compute (dy/dx)|x=0.
can anyone help me and shows the answer step by step? It is because i m a maths newbe.
dy/dx question
2007-10-14 9:29 am
回答 (2)
2007-10-14 6:00 pm
參考: Myself~~~
2007-10-14 10:11 am
Differentiate the equation.
2yln(x+1) y' + y^2/(x+1) -y'/(e^x+sinx) +y(e^x+cosx)/(e^x+sinx)^2 - 2sinx ln(x^2+1)+4xcosx/(x^2+1)=0
Sub x = 0
y^2 - y' +2y = 0
y' = y^2 + 2y
2yln(x+1) y' + y^2/(x+1) -y'/(e^x+sinx) +y(e^x+cosx)/(e^x+sinx)^2 - 2sinx ln(x^2+1)+4xcosx/(x^2+1)=0
Sub x = 0
y^2 - y' +2y = 0
y' = y^2 + 2y
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