Answer :
F(x)= -ln(-x3+3x-1) is a particular solution to the differential equation that passes through (1, 0). An equation for the tangent line to the graph of f at the point (1,0) = -3.
b.[tex]\frac{dy}{e^y} = (3x^2-6x)dx\\[/tex]
∫[tex]e^{-y} dy =[/tex] ∫[tex](3x^2-6x)dx[/tex]
[tex]-1e^{-y} = x^3-3x^2+C\\e^{-y} = -x^3+3x^2+C_{2}\\ -y= ln(-x^3+3x^2+C_{2}) \\y=-ln(-x^3+3x^2+C_{2}[/tex]
Given (1,0)
[tex]0= -ln (-1+3+C)\\0=ln_{c} (2+C)\\e^{0} = 2+C\\-1=C\\so f(x)= -ln(-x^3+3x^2-1)\\[/tex]
f(1.2)=-.465 as the actual value.
c. tangent line slope at (1,0).[tex]= e^0 (3(1)^2-6(1))\\[/tex]
[tex]=-3[/tex]
Line equation: y= 3(x-1) or y= 3x + 3
with f(1,2)=-3(1.2)+3 =-.6
To learn more about the differential equations, refer:-
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