dy = 8e^(4x)(1+y^2) dx
dy/(1+y^2) = 8e^(4x) dx
arctan(y) = 2e^(4x)+c
y = tan(2e^(4x)+c)
at x=0 you have
tan(2+c) = -2
c+2 = arctan(-2) = -1.107
c = -3.107
y = tan(2e^(4x)-3.107)
y′=8e^(4x)(1+y2)solve for y(0)=−2
⇒ y=
I tried to solve the problem and got the equation: tan^-1((2e^(4x)-2+tan(-2)))it says I am wrong and I resolved and got the same thing I must be doing something wrong please help!
2 answers
awesome! thank you so much! I see where I went wrong! Would it be ok if I asked you another one I am having trouble with?
1 answer