What is the general solution of the differential equation y ′ - 2 x y = x 3 ?

Name: What is the general solution of the differential equation y ′ - 2 x y = x 3 ?
Uploaded: 2022-02-23T17:22:57+00:00
Description: What is the general solution of the differential equation y ′ - 2 x y = x 3 ?

Question

What is the general solution of the differential equation       y    ′    -    2    x    y    =          x      3      ?

Katelyn Ryan · Accepted Answer

We have:      y    ′    -    2    x    y    =          x      3       ..... [1]This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;                    d        y                    d        x              +    P          (      x      )        y    =    Q          (      x      )      We can readily generate an integrating factor when we have an equation of this form, given by;      I    =          e              ∫        P                  (          x          )                          d          x                                    =          exp              (        ∫                 -        2        x                           d          x                )                            =          exp              (        -                  x          2                )                            =          e              -                  x          2                    And if we multiply the DE [1] by this Integrating Factor, I, we will have a perfect product differential;            e              -                  x          2                      y    ′    -    2    x          e              -                  x          2                      y    =          x      3              e              -                  x          2                          ∴          d              d        x                    (              e                  -                      x            2                              y      )        =          x      3              e              -                  x          2                    We can now integrate to get:            e              -                  x          2                      y    =    ∫               x      3              e              -                  x          2                                 d      x        +    C  The RHS integral can be evaluated by an application of Integration By Parts (omitted) which gives us:      ∫               x      3              e              -                  x          2                                 d      x        =    -          1      2              (              x        2            +      1      )              e              -                  x          2                    So we have:            e              -                  x          2                      y    =    -          1      2              (              x        2            +      1      )              e              -                  x          2                      +    C  Using the initial condition y(0)=2 we can evaluate the constant C      2          e      0        =    -          1      2              (      0      +      1      )              e      0        +    C    ⇒    C    =          5      2      Leading to the Particular Solution:            e              -                  x          2                      y    =    -          1      2              (              x        2            +      1      )              e              -                  x          2                      +          5      2            ∴    y    =    -          1      2              (              x        2            +      1      )              e              -                  x          2                      +          5              2                  e                      -                          x              2                                                                    =    -          1      2              (              x        2            +      1      )        +          5      2              e                        x          2                                              =          5      2              e                        x          2                      -          1      2              x      2        -          1      2

What is the general solution of the differential equation y'−2xy=x^3?

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