Question

solve this true or false...is it true or false?? justify your answer

Answer 1

answers:     i). True     ii). false   iii) true  iv) false   v)  true
==========================================
1. Let y1(x), y2(x) be two solutions.   Let  Y(x) = y2(x) - y1(x)
    Given  y1(0) = y2(0) = 0 .  Hence,  Y(0) = 0
               y1' = x² + y1²        , y2' = x²+y2²
    So Y' = y2' - y1' = (y2-y1) (y2+y1)
    =>  Y'(0) = Y(0) *(y2+y1) = 0

    Find Y'' = y2'' - y1'' = 2 (y2-y1)(x²+y2²+y1²-y1 y2)
    So    Y''(0) = Y(0) * (factor) = 0   

    Thus all derivatives of Y(x) at x=0 are zero, as Y(x) is a factor.
    Hence,  n th derivatives of y1 and y2 are always equal, in the
       neighbourhood of x=0. ie.,  - h< x < h, for some small h.

    Hence, if we take the Taylor series for y1 and y2, they will be equal.
    So there is only one unique solution for the given ODE. 
========
2.  Ortho Traj is not  x² + 2 y² = c²
     Given Parabolas :   y² = 4 a x
        y' = 4a/2y = y/2x   by eliminating a.
     Ortho traj :   y' = -2x/y
      So    y y' = - 2x
             y²/2 = - x² + c
============
3.  first find normal form of    y'' - 4 x y' + (4x²-1) y = 0
      y = u(x) v(x).                  here   p(x) = -4x     q(x) = 4x²-1 
                                                       p'(x) = -2x

     For normal form:  v u'' + Q(x) u = 0,  we find v, u and Q(x).
           v = exp(- integral p/2 dx) = exp(x²)
          Q(x) = p v' + v'' + q v = v (q - p²/4 - p'/2) 
                   = exp(x²) [4x² - 1 - 4x² - (-4/2)] = exp(x²) = v

    Normal form:  exp(x²) u'' + exp(x²) u = 0
    Thus the normal form for the given y'' - 4x y' + (4x²-1) y = -3 exp(x²) sin2x
      will be   u'' + u = -3 sin2x.... Just substitute and verify.
    In the question  the normal form is given in terms of v. Here I did in terms of u. u and v are exchanged.
===============
4)  given z = - [y + f(x-y)]
              partial derivatives:   dz/dx = - f '(x-y) ....  dz/dy = - 1 - f(x-y) * -1
    z² = y² + [f(x-y)]² + 2y f(x-y)
Differentiate (partial)
   2z dz/dx = 2 f(x-y) * f '(x-y) + 2 y f '(x-y)
   2z dz/dy = 2 y + 2 f(x-y) * f '(x-y) *(-1) + 2 f(x-y) + 2y f '(x-y)*(-1)
ADD these two eq:
     2 z [dz/dx + dz/dy ] = 2 y + 2 f(x-y) = - 2 z  (as given)
     so   dz/dx + dz/dy = -1   CONTRADICTION
===========
5.
  General Differential (partial DE) equation in two variables or 2nd degree:
    A u_xx + 2 B u_xy + C u_yy + D u_x + E u_y + F = 0 

    Evaluate  Z = Det | A   B |    =  AC - B².   If  Z < 0, then it is hyperbolic in x, y.
                                  | B   C | 
    For the given one:   A = 1 ,  B = x²/2 ,  C = -x²/2 - 1/4
    So  Z = - x²/2 - 1/4 - x⁴/4  < 0  for all x.
    So the solution   u(x,y) is hyperbolic.   u(x,y) = f(u_x, u_y, x, y)
    the solution can be simplified in the above manner.

Answer 2

$\huge\underline{\underline{\mathfrak{\red{A}\green{n}\pink{s}\orange{w}\blue{e}{r}}}}$

1. TRUE

2. FALSE

3. TRUE

4. FALSE

5. TRUE

____________________________________

20 thanks ❤️ = Inbox

Snapchat : vidhya_dawa2020

follow me ✨✨✨✨

Instagram : @buzy_.bee_