Stefan's Law of Radiation states that the radiation energy of abody is proportional to the fourth power of the absolute temperature T of a body. The rate of change of this energy ina surrounding medium of absolute temperature M is thus d T d t = k ( M 4 − T 4 )where k &gt; 0 is a constant. Show that the general solution of Stefan's equation is ln ⁡ | M + T M − T | + 2 tan − 1 ⁡ ( T M ) = 4 M 3 k t + cwhere c is an arbitrary constant.

Question

Stefan&#039;s Law of Radiation states that the radiation energy of abody is proportional to the fourth power of the absolute temperature T of a body. The rate of change of this energy ina surrounding medium of absolute temperature M is thus             d      T              d      t        =  k  (      M          4        −      T          4        )where k &amp;gt; 0 is a constant. Show that the general solution of Stefan&#039;s equation is  ln  ⁡      |              M      +      T              M      −      T            |    +  2      tan          −      1        ⁡  (      T    M    )  =  4      M          3        k  t  +  cwhere c is an arbitrary constant.

Lillianna Mendoza · Accepted Answer

d      T              d      t        =  k  (      M          4        −      T          4        )  ⇒            d      T                      M                  4                    −              T                  4                      k  d  t  ⇒  ∫      1          (              M                  2                    +              T                  2                    )      (              M                  2                    −              T                  2                    )        d  T  =  ∫  k  d  t  ⇒      1          M              4              ∫      1          (      1      +                        T                      2                                    M                      2                              )      (      1      −                        T                      2                                    M                      2                              )        d  T  =  ∫  k  d  tWe have,      1          (      1      +      n      )      (      1      −      n      )        =      1    2    ×            (      1      +      n      )      (      1      −      n      )              (      1      +      n      )      (      1      −      n      )        =      1    2    [      1          1      −      n        +      1          1      +      n        ]     w  h  e  r  e     n  =            T              2                    M              2            Our equation becomes      1          2              M                  4                      ∫  [      1          1      −                        T                      2                                    M                      2                                +      1          1      +                        T                      2                                    M                      2                                ]  d  T  =  ∫  k  d  t  ⇒      1    2    ∫  (      1                  M                  2                    −              T                  2                      +      1                  M                  2                    +              T                  2                      )  d  T  =  ∫  k  d  tIntegrating both sides, we shall get

Answered question

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New Questions in Calculus 1