How do I use mathematica to implicitly solve differential equations of a single variable?

Question

I'm trying to force Mathematica to implicitly differentiate an ellipse equation of the form:

x^2/a^2+y^2/b^2 == 100

with a = 8 and b = 6.

The command I'm using looks like this:

D[x^2/a^2 + y^2/b^2 == 100/. y -> 3/4*Sqrt[6400-x^2], x]

where, y->3/4*Sqrt[6400-x^2] comes from solving y in terms of x.

I got this far by following the advice found here: http://www.hostsrv.com/webmaa/app1/MSP/webm1010/implicit

Input for this script is the conventional way that an implicit relationship beween x and y is expressed in calculus textbooks. In Mathematica you need to make this relationship explicit by using y[x] in place of y. This is done automatically in the script by replacing all occurances of y with y[x].

But the solution Mathematica gives does not have y' or dy/dx in it (like when I solved it by hand). So I don't think it's been solved correctly. Any idea on what command would get the program to solve an implicit differential? Thanks.

Answer 1

The conceptually easiest option (as you mentioned) is to make y a function of x and use the partial derivative operator D[]

In[1]:= D[x^2/a^2 + y[x]^2/b^2 == 100, x]
        Solve[%, y'[x]]

Out[1]= (2 x)/a^2 + (2 y[x] y'[x])/b^2 == 0

Out[2]= {{y'[x] -> -((b^2 x)/(a^2 y[x]))}}

But for more complicated relations, it's best to use the total derivative operator Dt[]

In[3]:= SetOptions[Dt, Constants -> {a, b}];

In[4]:= Dt[x^2/a^2 + y^2/b^2 == 100, x]
        Solve[%, Dt[y, x]]

Out[4]= (2 x)/a^2 + (2 y Dt[y, x, Constants -> {a, b}])/b^2 == 0

Out[5]= {{Dt[y, x, Constants -> {a, b}] -> -((b^2 x)/(a^2 y))}}

Note that it might be neater to use SetAttributes[{a, b}, Constant] instead of the SetOptions[Dt, Constants -> {a, b}] command... Then the Dt doesn't carry around all that extra junk.

The final option (that you also mentioned) is to solve the original equation for y[x], although this is not always possible...

In[6]:= rep = Solve[x^2/a^2 + y^2/b^2 == 100, y]

Out[6]= {{y -> -((b Sqrt[100 a^2 - x^2])/a)}, {y -> (b Sqrt[100 a^2 - x^2])/a}}

And you can check that it satisfies the differential equation we derived above for both solutions

In[7]:= D[y /. rep[[1]], x] == -((b^2 x)/(a^2 y)) /. rep[[1]]

Out[7]= True

You can substitute your values a = 8 and b = 6 anytime with replacement rule {a->8, b->6}.

If you actually solve your differential equation y'[x] == -((b^2 x)/(a^2 y[x]) using DSolve with the correct initial condition (derived from the original ellipse equation) then you'll recover the solution for y in terms of x given above.