Question

Runge kutta method
X=0.1
dy/dx=x+y,x(0)=1

Answer 1

Answer:

(x,y) = (0.1,0.9)

Step-by-step explanation:

$\frac{dy}{dx} =x+y\\\frac{dy}{dx} -y =x$     Comparing this with $\frac{dy}{dx} +P y = Q(x)$

Here, P = -1   and Q(x) =x

Finding integral factor, I.F =$e^{\int Pdx}$

                                           = $e^{\int (-1)dx}$

                                      I.F = $e^{-x}$

Now, the solution of the differential equation is      

                                       $y(I.F) =\int {(I.F)Q(x)} \, dx +c$   [ c is integral constant]

    So,                              $y(e^{-x}) = \int{(e^{-1})(x) } , dx +c$

                                       $\frac{y}{e^{x} }= (-x e^{-x}+e^{-x} ) +c$

                                      $y = (1-x)+ce^{x}$

And    (x,y) =(1,0)  

So,    $0= (1-1) +ce^{1}$

        c = 0.

Now,   y =  1 - x

Therefore,  for x = 0.1,  y = 1-0.1= 0.9