Solve the given initial-value problem. (Use x as the independent variable.) 2y'y'' = 1, y(0) = 2, y'(0) = 1

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.

what is all that apply?

false

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