Question

Let y=f(x) be a particular solution to the differential equation dy/dx=xy^3 with f(1)=2. Write an equation for the tangent line to the graph of y=f(x) at x=1.

Answer 1

We're given the differential equation\[\frac{ dy }{ dx }=xy ^{3}.\] Recognize that this is a "separable d. e."?

Separating,

\[\frac{ dy }{ y ^{3} }=x*dx.\]
Now integrate both sides. The integral of the left side is \[\int\limits_{}^{}\frac{ dy }{ y ^{3} }=\int\limits_{}^{}y ^{-3}dy=\frac{ y ^{-2} }{-2 }.\]

Answer 2

Integrate the right side, remembering the constant of integration, C.

Apply the given initial condition to determine the value of C.

Answer 3

As it turns out, this is a trick question in that all this integration is unnecessary!!

To write the equation of the tangent line to the curve where x=1, we need two things:

(1) the slope of the tangent line. That's dy/dx, and its formula is given.
(2) the point at which the tangent line is tangent to the graph. That's given by f(1)=2: (1,2).

The slope of the tangent line at (1,2), from the given derivative, is m=(1)(2)^3, or 8.
The point of tangency is (1,2).
Use the point-slope formula to find the equation of this tangent line.