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[tex] y'(x)g(x)=0 [/tex] and [tex] y'(x)g(x)=\delta (x-a) [/tex]

except when x=a the two equations are equal , however the solutions are very different

[tex] y(x)=C [/tex] and [tex] y(x)= C+ \int dx \frac{\delta (x-a)}{g(x)} [/tex]

or using the properties of Dirac delta [tex] y(x)=C+\frac{1}{g(a)} [/tex]

the second equation depends on the form of g(x) whereas the first does not, however except at the point x=a the 2 ODE's are completely equal.