Question

solve y'=x+y,y(1)=0,find y(1,1) by Taylor's series​

Answer 1

Answer:

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Answer 2

Answer:

Taylor's Series method

Consider the one dimensional initial value problem

y' = f(x, y),   y(x0) = y0where 

f is a function of two variables x and y and (x0 , y0) is a known point on the solution curve.

If the existence of all higher order partial derivatives is assumed for y at x = x0, then by Taylor series the value of y at any neibhouring point x+h can be written as 

y(x0+h) = y(x0) + h y'(x0) + h2 /2 y''(x0) + h3/3! y'''(x0) +  .  .  .  .  .  .

where ' represents the derivative with respect to x.  Since at x0, y0 is known,  y' at x0 can be found by computing f(x0,y0).  Similarly higher derivatives of y at x0 also can be computed by making use of the relation  y'  = f(x,y) 

      y''  = fx + fyy'

      y''' = fxx + 2fxyy' + fyy y'2 + fyy''

 

and so on.   Then 

y(x0+h) = y(x0) + h f + h2  ( fx + fyy' ) / 2! + h3 ( fxx + 2fxyy' + fyy y'2 + fyy'' ) / 3! + o(h4)

Hence the value of  y  at any neighboring point  x0+ h  can be obtained by summing  the above infinite series.   However,  in any practical computation, the summation has to be terminated after some finite number of terms.  If the series has been terminated after the pthderivative term then the approximated formula is called the Taylor series approximation to y of order p and the error is of order p+1.  The same can be repeated to obtain y at other points of x in the interval [x0, xn] in a marching process. 

Step-by-step explanation:

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