The theorem of existence and uniqueness is: Let y ′ + p ( x ) y = g ( x ) be a first order linear differential equation such that p(x) and g(x) are both continuous for a &lt; x &lt; b. Then there is a unique solution that satisfies it.When a differential equation has no solution that satisfies y ( x 0 ) = y 0 , what does this mean?? Can the theorem be verified??

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The theorem of existence and uniqueness is: Let       y    ′    +  p  (  x  )  y  =  g  (  x  ) be a first order linear differential equation such that p(x) and g(x) are both continuous for   a  &amp;lt;  x  &amp;lt;  b. Then there is a unique solution that satisfies it.When a differential equation has no solution that satisfies   y  (      x          0        )  =      y          0      , what does this mean?? Can the theorem be verified??

scipionhi · Accepted Answer

The existence and uniqueness theorem for first-order linear differential equations can be stated as follows. Suppose that P and Q are continuous on the open interval I. If a and b are any real numbers, then there is a unique function y=f(x) satisfying the initial-value problem       y    ′    +  P  (  x  )  y  =  Q  (  x  ) with f(a)=b on the interval I. With regard to your question, the important point is that a and b are arbitrary real numbers and the unique solution f to the differential equation satisfies f(a)=b for every choice of a and b. Since every first-order linear differential equation satisfying the constraints of the theorem has a solution satisfying f(a)=b, there is no case in which such an equation has no solution satisfying f(a)=b.If we look at the simpler case of homogeneous first-order linear differential equations of the form       y    ′    +  P  (  x  )  y  =  0, where P is continuous on the open interval I, we can directly verify that for every choice of a and b, the function   f  (  x  )  =  b      e          −      A      (      x      )       where   A  (  x  )  =      ∫          a        x    P  (  t  )  d  t is a solution to       y    ′    +  P  (  x  )  y  =  0. Now letting g be an arbitrary solution of       y    ′    +  P  (  x  )  y  =  0, we establish uniqueness by showing that   g  (  x  )      e          A      (      x      )        =  b. Differentiating, we see that   h  (  x  )  =  g  (  x  )      e          A      (      x      )       is constant on the interval I. But h(a)=b, so we must have h(x)=b. This demonstrates that g=f. Notice that the choice of a and b does not affect the existence or uniqueness of solutions. Can you verify this in the case of non-homogeneous first-order linear differential equations of the form       y    ′    +  P  (  x  )  y  =  Q  (  x  )

The theorem of existence and uniqueness is: Let y ′ </msup> + p (

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