Question

The initial value problem dy/dx = x² + y² , y(0) = 0 has a unique solution in some interval of the form -h < x < h. is it true or false? justify your answer.

Answer 1

Yes. It is true.

See the diagram. That is the graph of  y(x)  for which:
          y '(x) = x² + y²,      y(x=0) = 0

Solution:
   y(x) = x³/3 + x⁷/(3²*7) + 2 x¹¹/(3²*7*11*3)
               + 13 x¹⁵/(3²*7*11*3*15*49) + 166 x¹⁹ /(3²*7*11*3*19*15*49*3)+..

We will find the solution by iteration methods.  We cannot express the solution to the given ODE in terms of known elementary functions. 
 
  One form of exact solution uses Bessel functions of first kind J_α (x) and second kind Y_α (x) with the order α = 1/4 and argument of x^2/2.  The Bessel function uses Gamma function with α = 1/4.

 An exact solution for the given problem may be given as a polynomial in terms of exp(x^3/3) and x, under Integral sign.

We can use Taylor's series expansion method or Picardi's method to find the solution as a polynomial in x , by using iterations.
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y ' = f(x,y) = x² + y² ,   y(0) = 0

$y_{k+1}(x)=y(0)+\int \limits_{0}^{x} {f(t,y_k)} \, dt=0+\int\limits_{0}^{x} {(t^2+y_k^2)} \, dt\\\\So\ Initial\ solution: \ y_1(x)=\frac{x^3}{3}\\\\y_2(x)=\frac{x^3}{3}+\int\limits_{0}^{x} {(t^3/3)^2} \, dt=\frac{x^3}{3}+\frac{x^7}{3^2*7}\\\\y_3(x)=\frac{x^3}{3}+\int\limits_{0}^{x} {(\frac{t^3}{3}+\frac{t^7}{3^2*7})^2} \, dt=\frac{x^3}{3}+\frac{x^7}{3^2*7}+\frac{2x^{11}}{3^2*7*11*3}+\frac{x^{15}}{63^2*15}\\\\y_4(x)=\frac{x^3}{3}+\int \limits_{0}^{x} {(y_3(t))^2} \, dt$

Thus
y₄(x) = x³/3 + x⁷/(3²*7) + 2 x¹¹/(3²*7*11*3)
               + 13 x¹⁵/(3²*7*11*3*15*49) + 166 x¹⁹ /(3²*7*11*3*19*15*49*3)+..

So we see that we have a unique solution for  - h < x < h  in the neighbourhood of zero.