System of Differential Equations for Particular Initial Conditions { x ′ ( t ) = x ( t ) + y ( t ) y ′ ( t ) = y ( t ) With the intial conditions x ( 0 ) = 1 and y ( 0 ) = 1.Obviously, y ( t ) (as well as y ′ ( t ) are equal to e t .How do I find x ( t ) now?I have tried the following: x ′ ( t ) = x ( t ) + y ( t ) can be written as y ( t ) = x ′ ( t ) − x ( t ) .Now we know that x ( 0 ) and y ( 0 ) = 1, hence: 1 = x ′ ( 0 ) − 1 , ∴ x ′ ( 0 ) = 2..How do I proceed from here?

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System of Differential Equations for Particular Initial Conditions      {                                        x            ′                    (          t          )          =          x          (          t          )          +          y          (          t          )                                                  y            ′                    (          t          )          =          y          (          t          )                        With the intial conditions   x  (  0  )  =  1 and   y  (  0  )  =  1.Obviously,   y  (  t  ) (as well as       y    ′    (  t  ) are equal to       e    t  .How do I find   x  (  t  ) now?I have tried the following:      x    ′    (  t  )  =  x  (  t  )  +  y  (  t  ) can be written as  y  (  t  )  =      x    ′    (  t  )  −  x  (  t  )  .Now we know that   x  (  0  ) and   y  (  0  )  =  1, hence:   1  =      x    ′    (  0  )  −  1  ,  ∴      x    ′    (  0  )  =  2..How do I proceed from here?

xxsailojaixxv5 · Accepted Answer

Since you&#039;ve found that   y  (  t  )  =      e    t  , then you can plug it in to the formula for   x, in order to get that       x    ′    (  t  )  =  x  (  t  )  +      e    t  , which is a linear ODE, and therefore, you can begin by solving the homogenous ODE       x    ′    (  t  )  =  x  (  t  ), and then use variation or some other method to solve the non-homogenous ODE.

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