In applications, most inflal value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial walue problem (dy)/(dx)=f(x,y),y(x_(0))=y_(0). If f and (df)/(dy) are continuous functions in some rectangle x_(0),y_(0) ph solutionspile.com

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In applications, most inflal value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial walue problem

(dy)/(dx)=f(x,y),y(x_(0))=y_(0)

. If

f

and

(df)/(dy)

are continuous functions in some rectangle

x_(0),y_(0)\phi (x)(dy)/(dx)=y^((1)/(3))(dy)/(dx)=y^((1)/(3))y^(-(1)/(3))dy=1dx?x_(0)-5, where 5 is a positive number. The method for separable equations can give a solution, but it may not give all the solutions. To Rustrate this, consider the equation (dy)/(dx)=y^((1)/(3)). Answer parts (a) through (d).
(a) Use the method of secaration of variables to find the solution to (dy)/(dx)=y^((1)/(3)). Begin by separating the variabies. y^(-(1)/(3))dy=1dx
Solve the differential equation, ignoring lost solutions, fary.
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