Question
Consider the differential equation \(\frac{dy}{dx}=\frac{y^{2}}{x-1}.\)
(a) On the axes provided, sketch a slope field for the given differential equation at the six points indicated.
(b) Let y = f(x) be the particular solution to the given differential equation with the initial condition f(x) = 3. Write an equation for the line tangent to the graph of y=f(x) at x= 2.
Use your equation to approximate f(2.1).
(c) Find the particular solution y = f (x) to the given differential equation with the initial condition f (2) = 3.
Answer/Explanation
Ans:
(a) 
(b) \(\frac{dy}{dx}_{(x,y)=(2,3)}=\frac{3^{2}}{2-1}=9\)
An equation for the tangent line is y = 9(x – 2) + 3.
f(2.1) ≈ 9(2.1-2) + 3 = 3.9
(c) \(\frac{1}{y^{2}}dy=\frac{1}{x-1}dx\)
\(\int \frac{1}{y^{2}}dy=\int \frac{1}{x-1}dx\)
\(-\frac{1}{y}= ln|x-1|+C\)
\(-\frac{1}{3}= ln|2-1|+C\Rightarrow C=-\frac{1}{3}\)
\(-\frac{1}{y}= ln|x-1|-\frac{1}{3}\)
\(y= \frac{1}{\frac{1}{3}-ln(x-1)}\)
Note: This solution is valid for 1 < x < 1 + e1/3.