I would need some help in solving the following differential equation: u ′ ( p ) + r + N λ p N λ p ( 1 − p ) u ( p ) = r + N λ p N λ ( 1 − p ) g .I already know that the solution is of the form g p + C ( 1 − p ) ( 1 − p p ) r / λ N ,,where C is a constant.Now this is a standard linear first-order differential equation. Letting f ( p ) = r + N λ p N λ p ( 1 − p ) , q ( p ) = r + N λ p N λ ( 1 − p ) g , μ ( p ) = e ∫ f ( p ) d p ,the solution should have the form u ( p ) = C μ ( p ) + 1 μ ( p ) ∫ μ ( p ) q ( p ) d p ,where C is a constant.The calculation of μ ( p ) was not much of a problem. Indeed, we can rewrite f(p) as f ( p ) = r N λ 1 p + ( 1 + r N λ ) 1 1 − p Using ∫ 1 p d p = ln ⁡ ( p ) and ∫ 1 1 − p d p = − ln ⁡ ( 1 − p ), I found that μ ( p ) = p r N λ ( 1 − p ) − 1 − r N λ .Where I have difficulties is calculating the term ∫μ(p)q(p)dp, where μ ( p ) q ( p ) = r g N λ p r N λ ( 1 − p ) − 2 − r N λ + p r N λ + 1 ( 1 − p ) − 2 − r N λ .How do I integrate this last term? Or did I miss something obvious, because in the paper I'm looking at they don't provide any details about how they get to the solution of the differential equationEdit: small typo corrected in the last displayed equation.Edit: there is a big typo in the differential equation: the rhs should have r + N λ rather than r + N λ p in the numerator. This probably makes it much simpler to solve. My bad.

Question

I would need some help in solving the following differential equation:      u    ′    (  p  )  +            r      +      N      λ      p              N      λ      p      (      1      −      p      )        u  (  p  )  =            r      +      N      λ      p              N      λ      (      1      −      p      )        g  .I already know that the solution is of the form  g  p  +  C  (  1  −  p  )            (                        1          −          p                p            )              r              /            λ      N        ,,where C is a constant.Now this is a standard linear first-order differential equation. Letting  f  (  p  )  =            r      +      N      λ      p              N      λ      p      (      1      −      p      )        ,        q  (  p  )  =            r      +      N      λ      p              N      λ      (      1      −      p      )        g  ,        μ  (  p  )  =      e          ∫      f      (      p      )      d      p        ,the solution should have the form  u  (  p  )  =      C          μ      (      p      )        +      1          μ      (      p      )        ∫  μ  (  p  )  q  (  p  )  d  p  ,where C is a constant.The calculation of   μ  (  p  ) was not much of a problem. Indeed, we can rewrite f(p) as  f  (  p  )  =      r          N      λ            1    p    +      (    1    +          r              N        λ              )        1          1      −      p      Using   ∫      1    p    d  p  =  ln  ⁡  (  p  ) and   ∫      1          1      −      p        d  p  =  −  ln  ⁡  (  1  −  p  ), I found that  μ  (  p  )  =      p                  r                  N          λ                      (  1  −  p      )          −      1      −              r                  N          λ                      .Where I have difficulties is calculating the term ∫μ(p)q(p)dp, where  μ  (  p  )  q  (  p  )  =            r      g              N      λ            p                  r                  N          λ                      (  1  −  p      )          −      2      −              r                  N          λ                      +      p                  r                  N          λ                    +      1        (  1  −  p      )          −      2      −              r                  N          λ                      .How do I integrate this last term? Or did I miss something obvious, because in the paper I&#039;m looking at they don&#039;t provide any details about how they get to the solution of the differential equationEdit: small typo corrected in the last displayed equation.Edit: there is a big typo in the differential equation: the rhs should have   r  +  N  λ rather than   r  +  N  λ  p in the numerator. This probably makes it much simpler to solve. My bad.

elladanzaez · Accepted Answer

After the OP corrected the typo, the PDE becomes easy to solve      u    ′    (  p  )  +            r      +      N      λ      p              N      λ      p      (      1      −      p      )        u  (  p  )  =            r      +      N      λ              N      λ      (      1      −      p      )        g  a  =      r          N      λ            u    ′    (  p  )  +            a      +      p              p      (      1      −      p      )        u  (  p  )  =            a      +      1              (      1      −      p      )        gAn obvious particular solution is u=gp. Thus, change of function : u(p)=v(p)+gp      v    ′    +  g  +            a      +      p              p      (      1      −      p      )        (  v  +  g  p  )  =            a      +      1              (      1      −      p      )        gAfter simplification :      v    ′    +            a      +      p              p      (      1      −      p      )        v  =  0Separable ODE easy to solve :  v  =  C            (      1      −      p              )                  1          +          a                            p      a        u  =  g  p  +  C            (      1      −      p              )                  1          +          a                            p      a

I would need some help in solving the following differential equation: u ′

Answered question

Answer & Explanation

New Questions in Differential Equations