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A singular solution y_{s}(x) of an ordinary differential equation is a solution that is tangent to every solution from the family of general solutions. By tangent we mean that there is a point x where y_{s}(x) = y_{c}(x) and y'_{s}(x) = y'_{c}(x) where y_{c} is any general solution.## Example

Consider the following Clairaut's equation:## See also

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.

- $y(x)\; =\; x\; cdot\; y\text{'}\; +\; (y\text{'})^2\; ,!$

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:

- $p\; =\; y\text{'}\; =\; p\; +\; x\; p\text{'}\; +\; 2\; p\; p\text{'}\; ,!$

which by simple algebra yields

- $0\; =\; (2\; p\; +\; x\; )p\text{'}.\; ,!$

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 + 2p = 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 is a singular solution.

First condition of tangency: y_{s}(x) = y_{c}(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 ($-2c\; ,\; -c^2$).

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

We calculate the derivatives:

- $y\_c\text{'}(-2\; cdot\; c)\; =\; c\; ,!$

- $y\_s\text{'}(-2\; cdot\; c)\; =\; -(1/2)\; cdot\; x\; |\_\{x\; =\; -2\; cdot\; c\}\; =\; c.\; ,!$

We see that both requirements are satisfied and therefore y_{s} is tangent to general solution y_{c}. Hence,

- $y\_s(x)\; =\; -(1/4)\; cdot\; x^2\; ,!$

is a singular solution for the family of general solutions

- $y\_c(x)\; =\; c\; cdot\; x\; +\; c^2\; ,!$

of this Clairaut equation:

- $y(x)\; =\; x\; cdot\; y\text{'}\; +\; (y\text{'})^2.\; ,!$

Note: The method shown here can be used as general algorithm to solve any Clairaut's equation, i.e. first order ODE of the form

- $y(x)\; =\; x\; cdot\; y\text{'}\; +\; f(y\text{'}).\; ,!$

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Last updated on Tuesday September 30, 2008 at 10:26:55 PDT (GMT -0700)

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This article is licensed under the GNU Free Documentation License.

Last updated on Tuesday September 30, 2008 at 10:26:55 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

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