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Question

Consider the differential equation dy/dx =x^2y^2with a particular solution y=f(x) having an initial condition y(−3) = 1 . Use the equation of the line tangent to the graph of f at the point (−3, 1) in order to approximate the value of f(−2.8) .

 

Answer 1

The dy/dx=x²y²,

dy/y²=x²dx,

Integrating:

-1/y=x³/3+C.

Plug in (-3,1):

-1=-9+C, C=8, -1/y=x³/3+8,

y=-3/(x³+24). Note that x=∛-24=-2.8845 approx is an asymptote, and -3 is on one side of the asymptote while -2.8 is on the other side, so the linear approximation (tangent line at (-3,1)) is not going to help to get an approximation for y(-2.8).

At (-3,1), dy/dx=(-3)²(1)²=9.

So y-1=9(x+3) is the linear approximation.

That is, y=9x+28.

When x=-2.8, y=-25.2+28=2.8.

(Compare with y(-2.8): y=-3/(-21.952+24)=-1.465 approx.)

Because of the large discrepancy between the approximate and actual values, I think that the question contains at least one error. Should dy/dx=x²/y², for example?