Answer: y(x) = (2/3)*(x + 1)^(3/2) + 1/3.
Step-by-step explanation:
We have the differential equation:
2*y'(x)*y''(x) = 1.
y(0) = 2
y'(0) = 1.
I will use the change:
u = y'
u'= y''
Then we must solve:
2*u*u' = 1.
The product does not depend on x:
u*u' = 1/2.
By looking at the problem, i know that the functions must be something like:
u = a*(x + b)^n
Where a and b are real numbers.
u' = n*a*(x + b)^(n - 1)
such that:
(x + b)^n*(x + b)^(n - 1) does not depend on x
This means that:
(x + b)^n*(x + b)^(n - 1) = (x + b)^(n + n - 1) = 0
n + n - 1 = 0
2n = 1
n = 1/2
Then:
u = a*(x + b)^(1/2)
u' = (a/2)*(x + b)^(-1/2)
The differential equation becomes:
u*u' = 1/2
a*(x + b)^(1/2)* (a/2)*(x + b)^(-1/2) = 1/2
(1/2)*a^2 = 1/2
a^2 = 1.
And by the initial conditions, we have:
y'(0) = u(0) = 1
then:
u(0) = a*(0 + b)^(1/2) = a*b^(1/2) = 1
now, if a = -1, then b^(1/2) must be negative, this can not really be then we must have:
a = 1
u(0) = 1*b^(1/2) = 1
then b = 1.
u(x) = (x + 1)^(1/2).
And remember that y'(x) = u(x).
Then we need to integrate u(x) over x.
let's use the change of variables:
w = x + 1
dw = dx
Then the integration of u(x) is:
y(w) = ∫(w)^(1/2)dw = (2/3)*w^(3/2) + c
where c is a constant of integration.
now we can go back to x:
y(x) = (2/3)*(x + 1)^(3/2) + c
And we know that:
y(0) = 1 = (2/3)*(0 + 1)^(3/2) + c
1 = (2/3)*1 + c
1 - 2/3 = c
1/3 = c
Then:
y(x) = (2/3)*(x + 1)^(3/2) + 1/3.
