So my course shows me three differential equations: x ˙ + x 2 = t x ˙ = ( t 2 + 1 ) ( x − 1 ) x ˙ + x = t 2 The first one is not a linear ordinary differential equation (ODE) apparently, the other two are.Unfortunately, they don't show a clear way how to find out if an ODE is linear or not. So how we can find out if an ODE is linear?For the second one, I thought I bring it into standard form somehow: x ˙ = ( t 2 + 1 ) ( x − 1 ) = x t 2 + x − t 2 − 1 = x ( t 2 + 1 ) − t 2 − 1If we say we let p ( t ) = t 2 + 1 and q ( t ) = 1 + t 2 , then we could say: x ˙ = x p ( t ) − q ( t ) = . . .And so on, to simplify until we reach standard form of a linear ODE (or not).Is that the way to go? Or is there some other way to check if a ODE is linear?

Question

So my course shows me three differential equations:            x      ˙        +      x    2    =  t            x      ˙        =  (      t    2    +  1  )  (  x  −  1  )            x      ˙        +  x  =      t    2  The first one is not a linear ordinary differential equation (ODE) apparently, the other two are.Unfortunately, they don&#039;t show a clear way how to find out if an ODE is linear or not. So how we can find out if an ODE is linear?For the second one, I thought I bring it into standard form somehow:            x      ˙        =  (      t    2    +  1  )  (  x  −  1  )  =  x      t    2    +  x  −      t    2    −  1  =  x  (      t    2    +  1  )  −      t    2    −  1If we say we let   p  (  t  )  =      t    2    +  1 and   q  (  t  )  =  1  +      t    2  , then we could say:            x      ˙        =  x  p  (  t  )  −  q  (  t  )  =  .  .  .And so on, to simplify until we reach standard form of a linear ODE (or not).Is that the way to go? Or is there some other way to check if a ODE is linear?

Haggar72 · Accepted Answer

Accordingly, the first is not linear, and the last two ode&#039;s are linear. Note that we just need that the equation is a linear polynomial in   x and its derivatives. That is, of the form:       a          n      +      1        (  t  )      x          (      n      )        +  ⋯  +      a    1    (  t  )  x  +      a    0    (  t  )  =  0. The functions       a    i    (  t  ) needn&#039;t be linear.

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