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Difference between unique solution and particular solution

by inter060708
Tags: difference, solution, unique
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Jan10-13, 12:43 AM
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For first order ordinary differential equations.

Are each particular solution a subset of unique solutions? Please explain, thanks
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Jan10-13, 09:59 AM
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A particular solution of an equation is a specific function that satisfies the equation.

e.g. y' = y. A particular solution is y = 2e^x.

Uniqueness is a word that means that there is only one. In the context of Intro to Diff. Eq. it is used in connection with initial value problems.

e.g. y'=y, y(0) = 5.
There is only one solution of this initial value problem: y=5e^x. We say that this problem has a unique solution.

Consider [itex]y'=(3/2)y^{1/3}, y(0)=0 [/itex].
y=0 (the constant function) is a solution of this problem. y=x^(3/2) is another solution of this problem. This problem does not have a unique solution.

Consider (y')^2+y^2=1 with initial value y(0)=1. Then y = 1 (the constant function) and y=cos(x) are two solutions. Therefore this problem also does not have a unique solution.
Jan11-13, 02:52 PM
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Quote Quote by inter060708 View Post
For first order ordinary differential equations.

Are each particular solution a subset of unique solutions? Please explain, thanks
It's the other way around. A "unique solution" to a differential equation with specific conditions is one of an infinite number of specific solutions to the equation. So the set of all solutions to a given differential equation with additional conditions is a subset, with one member, of the set of all "specific solutions".

For example, the differential equation y''= y has "general solution" [itex]y(x)= Ae^x+ Be^{-x}[/itex]. That is, any choice of values for A and B give a solution and every solution is of that form. If we choose specific values for A and B, say arbitrarily select A= 1, B= 2, we have [itex]y(x)= e^x+ 2e^{-x}[/itex] which is a "specific" solution.

If, instead, we have the "initial value problem" with differential equation y''= y and the conditions that y(0)= 1, y'(0)= 0, then we have a unique solution- it must be of the form [itex]y(x)= Ae^x+ Be^{-x}[/itex], as must all solutions, and we must have [itex]y(0)= Ae^0+ Be^0= A+ B= 1[/itex] and [itex]y'(0)= Ae^0- Be^0= A- B= 0[/itex]. Those two equations are satisfied by A= B= 1/2. That is, the unique solution to this problem is [itex]y(x)= (1/2)e^x+ (1/2)e^{-x}[/itex].

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