The problem reads:The solution of a certain differential equation is of the form y ( t ) = a exp ⁡ ( 5 t ) + b exp ⁡ ( 8 t )where a and b are constants.The solution has initial conditions y ( 0 ) = 5 and y ′ ( 0 ) = 5Find the solution by using the initial conditions to get linear equations for a and b. ....................What I did was solve using the initial conditions and I found that a + b = 5and 5 a + 8 b = 5.Am I totally on the wrong track? I don't know what it means to find a linear equation for a and b. I'd appreciate it if you could solve it step by step.

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The problem reads:The solution of a certain differential equation is of the form  y  (  t  )  =  a  exp  ⁡  (  5  t  )  +  b  exp  ⁡  (  8  t  )where   a and   b are constants.The solution has initial conditions   y  (  0  )  =  5 and       y    ′    (  0  )  =  5Find the solution by using the initial conditions to get linear equations for   a and   b. ....................What I did was solve using the initial conditions and I found that  a  +  b  =  5and   5  a  +  8  b  =  5.Am I totally on the wrong track? I don&#039;t know what it means to find a linear equation for   a and   b. I&#039;d appreciate it if you could solve it step by step.

Schetterai · Accepted Answer

You&#039;re absolutely right.If   y  (  t  )  =  a            e              5      t        +  b            e              8      t       then       y    ′    (  t  )  =  5  a            e              5      t        +  8  b            e              8      t      .Hence   y  (  0  )  =  a            e        0    +  b            e        0    =  a  +  b and       y    ′    (  0  )  =  5  a            e        0    +  8  b            e        0    =  5  a  +  8  b.To solve   y  (  0  )  =  5 and       y    ′    (  0  )  =  5 you need to solve   a  +  b  =  5 and   5  a  +  8  b  =  5 simultaneously.Personally I would use matrix algebra, but it&#039;s up to you.      (                            1                          1                                      5                          8                      )        (                            a                                      b                      )    =      (                            5                                      5                      )  The two-by-two matrix on the right has determinant   1  ×  8  −  5  ×  1  =  3  ≠  0 and so there is a unique solution. Multiplying the left and right by the inverse matrix gives      (                            a                                      b                      )    =      1    3        (                            8                          −          1                                      −          5                          1                      )        (                            5                                      5                      )  Expanding the right hand side gives                    a                            =                              1          3                (        8        ×        5        −        1        ×        5        )                            =                              35          3                                                  b                            =                              1          3                (        −        5        ×        5        +        1        ×        5        )                            =                    −                  20          3                    Your final solution is then                              y          (          t          )          =                                    35              3                                                          e                                      5              t                                −                                    20              3                                                          e                                      8              t

The problem reads: The solution of a certain differential equation is of the form y (

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