Solve the initial value problem yy x x2 y205 with y43305 a

Solve the initial value problem yy? + x = (x^2 +y^2)^0.5 with y(4)=(33)^0.5.

a) To solve this, we should use the substitution
u=
u ? =
Enter derivatives using prime notation (e.g., you would enter y ? for dy dx ).

b) After the substitution from the previous part, we obtain the following linear differential equation in x,u,u ? .

c) The solution to the original initial value problem is described by the following equation in x,y .

yy\' + x = (x^2 + y^2)^(0.5)

Let u = x^2 + y^2 --> First answer to A

So, u\' = 2x + 2yy\' --> Second answer to A

u\'/2 = x + yy\'

So, the integral becomes :

u\'/2 = u^(0.5) --> ANSWER to b

u\'/u^(0.5) = 2

u^(-0.5)*du = 2dx

Integrating :

2u^(0.5) = 2x + C

Plug in for u :

2*sqrt(x^2 + y^2) = 2x + C

sqrt(x^2 + y^2) = x + D

x^2 + y^2 = (x + D)^2

y^2 = (x + D)^2 - x^2

y = sqrt((x + D)^2 - x^2)

y(5) = sqrt(33)

sqrt(33) = sqrt((4 + D)^2 - 4^2)

33 = (4 + D)^2 - 4^2)

33 = 16 + 8D + D^2 - 16

D^2 + 8D - 33 = 0
(D + 11)(D - 3) = 0

D = -11 or 3

So, the solution is :

y = sqrt((x - 11)^2 - x^2)
y = sqrt(x^2 - 22x + 121 - x^2)
y = sqrt(121 - 22x) ---> FIRST ANSWER for c

y = sqrt((x + 3)^2 - x^2)
y =sqrt(x^2 + 6x + 9 - x^2)
y = sqrt(6x + 9) --> SECOND POSSIBLE ANSWER for c