Find a solution for a differential equation, assume that y ( x ) = a x 2 + b x + cFind a solution to the given equation: y ′ + 2 y = x 2 + 4 x + 7and we are told to assume that y(x) is a quadratic function, which follows this general form y = a x 2 + b x + c .Now to find the solutions, at first I tried to solve the given for y and then see what I could do from there but, I realized that you can't take the derivative of a function that contains its own derivative.So then I thought that I had to create my own function that follows the perviously mentioned parameters (the given and the quad. form). This is were I am stuck, I can not mathematically deduce a solution for this problem.

Question

Find a solution for a differential equation, assume that   y  (  x  )  =  a      x    2    +  b  x  +  cFind a solution to the given equation:      y    ′    +  2  y  =      x    2    +  4  x  +  7and we are told to assume that y(x) is a quadratic function, which follows this general form  y  =  a      x    2    +  b  x  +  c  .Now to find the solutions, at first I tried to solve the given for y and then see what I could do from there but, I realized that you can&#039;t take the derivative of a function that contains its own derivative.So then I thought that I had to create my own function that follows the perviously mentioned parameters (the given and the quad. form). This is were I am stuck, I can not mathematically deduce a solution for this problem.

Biardiask3zd · Accepted Answer

Step 1If   y  =  a      x    2    +  b  x  +  c, then       y    ′    =  2  a  x  +  b. Plug this into your equation:                              y          ′                +        2        y                            =        2        a        x        +        b        +        2        (        a                  x          2                +        b        x        +        c        )                                          =        2        a                  x          2                +        (        2        a        +        2        b        )        x        +        b        +        2        c                                                                              =                                      !                                                x          2                +        4        x        +        7.            Step 2Two polynomials are equal iff they are equal in all coefficients. You obtain a linear equation system      {                            2          a                          =          1                                      2          a          +          2          b                          =          4                                      b          +          2          c                          =          7                        which you have to solve for (a,b,c) in order to get a particular solution.

Paula Cameron · Accepted Answer

Step 1If you assume that   y  =  a      x    2    +  b  x  +  c, you can plug this hypothesis in the equation, which then writes  (  2  a  x  +  b  )  +  2  (  a      x    2    +  b  x  +  c  )  =      x    2    +  4  x  +  7.For the two members to be equal for all x, by regrouping the terms, you must have that   2  a  =  1, then   2  a  +  2  b  =  4 and finally   b  +  2  c  =  7.As these equations do have a solution, the hypothesis works.More generally, if your LHS is a linear combination of the derivatives of the unknown, and the RHS is a polynomial, you can try an arbitrary polynomial of the same degree.Step 2Also note that the solution that you found is not the only one which is possible. Indeed, you can add any function       y    ∗   such that                    y        ∗              ′    +  2      y    ∗    =  0without violating the equation. And       y    ∗    =  c      e          −      2      x       where c has any value is such a function.

Find a solution for a differential equation, assume that y(x)=ax^2+bx+c

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