Question

solve the linear differential equation (1+x)dy/dx-xy=1-x

Answer 1

Please see the attachment

Answer 2

The solution of the linear differential equation

(1 + x)dy/dx - xy = 1 - x is given by

$ye^{-x} (1 + x) = -xe^{-x} + C$

• We have ,

                (1 + x)dy/dx - xy = 1 - x

  Dividing the above equation by (1 + x), we get

       $\frac{dy}{dx} - \frac{x}{1 + x}y = \frac{1 - x}{1 + x}$

• Now this differential equation is in the form  

    dy/dx - y P(x) = Q(x)  

    where P(x) = $\frac{-x}{1 + x}$ and Q(x) =  $\frac{1 - x}{1 + x}$

• Now we will find the integrating factor (I.F)

  I.F = $e^{\int\limits {P(x)} \, dx }$

        = $e^{-\int\limits {\frac{x}{1 + x} } \, dx }$

        = $e^{-\int\limits {\frac{1 + x -1}{1 + x} } \, dx }$

        = $e^{-\int\limits {1 -\frac{1}{1 + x} } \, dx }$

        = $e^{-x + log(1+x)}$

        = $e^{-x} e^{log(1 + x)}$

    I.F = $(1 + x) e^{-x}$        -   (1)

• The solution is given by  

  y (I.F) = ∫ Q(x) (I.F) dx

  Now, putting the value of I.F from (1) in the above equation ,we get

  $ye^{-x} (1 + x) = \int\limits {e^{-x}(1 + x)\frac{1 - x}{1 + x} } \, dx$

  $ye^{-x} (1 + x) = \int\limits {e^{-x}(1 - x) } \, dx$    - (2)

• Solving RHS

$\int\limits {e^{-x}(1 - x) } \, dx$

Let -x = t

     -dx = dt

∴ $\int\limits {e^{t}(1 + t) } \, dt$

• Now it is in the form $\int\limits {e^{t}(f(t) + f'(t) } \, dt$

  where f(t) = t and f'(t) = 1

  it's integration is given by  

         $\int\limits {e^{t}(f(t) + f'(t) } \, dt = e^{t} f(t)$ + C

therefore,

        $\int\limits {e^{t}(1 + t) } \, dt = e^{t} t$ + C

  ∴ $\int\limits {e^{-x}(1 - x) } \, dx = -xe^{-x}$ + C     - (3)

• Equating (2) and (3) , we get

∴  $ye^{-x} (1 + x) = -xe^{-x} + C$

Or $y = \frac{-x}{1 + x} + C'$