Clairaut's Equation

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Clairaut's Equation

:

y(x)=xrac{dy}{dx}+f\left(rac{dy}{dx}
ight).

To solve such an equation, we differentiate with respect to ''x'', yielding

:

rac{dy}{dx}=rac{dy}{dx}+xrac{d^2 y}{dx^2}+f'\left(rac{dy}{dx}
ight)rac{d^2 y}{dx^2},

so

:

0=\left(x+f'\left(rac{dy}{dx}
ight)
ight)rac{d^2 y}{dx^2}.

Hence, either

:

0=rac{d^2 y}{dx^2}

or

:

0=x+f'\left(rac{dy}{dx}
ight).

In the former case, ''C'' = ''dy''/''dx'' for some constant ''C''. Substituting this into the Clairaut's equation, we have the family of functions given by

:

y(x)=Cx+f(C),\,

the so-called ''general solution'' of Clairaut's equation.

The latter case,

:

0=x+f'\left(rac{dy}{dx}
ight),

defines only one solution ''y''(''x''), the so-called '' Singular Solution '', whose graph is the Envelope of the graphs of the general solutions. The singular solution is usually represented using parametric notation, as (''x''(''p''), ''y''(''p'')), where ''p'' represents ''dy''/''dx''.