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Usually, singular solutions appear in differential equations when there is a need to divide in a term that might be equal to Zero . Therefore, when one is solving a differential equation and using division one must check what happens if the term is equal to zero, and whether it leads to a singular solution.


EXAMPLE


Consider the following Clairaut's Equation :

: y(x) = x \cdot y' + (y')^2 \,\!

where primes denote derivatives with respect to ''x''. We write ''y' = p'' and then

: y(x) = x \cdot p + (p)^2 \,\!

Now, we shall take the differential according to ''x'':

: p = y' = p + x p' + 2 p p' \,\!

which by simple Algebra yields

: 0 = ( 2 p + x )p' \,\!

This condition is solved if ''2p+x=0'' or if ''p'=0''.

If ''p' '' = 0 it means that ''y' = p = c'' = constant, and the general solution is:

: y_c(x) = c \cdot x + c^2 \,\!

where ''c'' is determined by the initial value.

If ''x'' + 2''p'' = 0 than we get that ''p'' = −(1/2)''x'' and substituting in the ODE gives

: y_s(x) = -(1/2)x^2 + (-(1/2)x)^2 = -(1/4) \cdot x^2 \,\!

Now we shall check whether this a singular solution.

First condition of tangency: ''ys''(''x'') = ''yc''(''x''). We solve

: c \cdot x + c^2 = y_c(x) = y_s(x) = -(1/4) \cdot x^2 \,\!

to find the intersection point, which is (−2''c'', −''c'').

Second condition tangency: ''y's''(''x'') = ''y'c''(''x'').

We calculate the Derivative s:

: y_c'(-2 \cdot c) = c \,\!