Question

What is the solution of the differential equation? : `ydy/dx = (9-4y^2)^(1/2)` where `x=0` when `y=0`

Answer 1

`y = +-sqrt(6x-4x^2)`

Explanation:

We have:

`ydy/dx = (9-4y^2)^(1/2)` ..... [A}

Which is a First Order separable (non-linear) Differential Equation, so we can rearrange [A] to get:

`y/sqrt(9-4y^2) dy/dx = 1`

And we can now "seperate the variables as follows":

`int \ y/sqrt(9-4y^2) \ dy = int \ dx` ..... [B]

Here the RHS is trivial to integrate, and we will need a substitution for the LHS integral:

`I = int \ y/sqrt(9-4y^2) \ dy`

Let us perform a substitution, and try:

`u = 4y^2 => (du)/dy = 8y`

Then substituting into the integral, we get:

`I = 1/8 \ int \ (8y)/sqrt(9-4y^2) \ dy`
`\ \ = 1/8 \ int \ (1)/sqrt(9-u) \ du`

And we now directly integrate this, giving:

`I = 1/8 (-2sqrt(9-u))`
`\ \ = -1/4 sqrt(9-u)`

And reversing the substitution we get:

`I = -1/4 sqrt(9-4y^2)`

Using this result, we can now integrate our earlier result [B] to get the General Solution:

`-1/4 sqrt(9-4y^2) = x + C`

We can apply the initial conditions `x=0` when `y=0` to fin `C`:

`-1/4 sqrt(9-0) = 0 + C => C =-3/4`

Hence, the Particular Solution is:

`-1/4 sqrt(9-4y^2) = x -3/4`

And for an explicit solution:

`sqrt(9-4y^2) = 3-4x`
`:. 9-4y^2 = (3-4x)^2` ..... `(star)`
`:. 9-4y^2 = 9-24x+16x^2`
`:. 4y^2 = 24x-16x^2`
`:. y^2 = 6x-4x^2`
`:. y = +-sqrt(6x-4x^2)`, confirming the given solution.

(NB, Care should be taken in interpreting the solution within the context of the model of the DE, as step `(star)` may introduce an invalid negative solution).

Answer 2

See below.

Explanation:

Making

`z = (2/3 y)^2` we get at the equivalent system

`9/8 z'- 3sqrt[1 - z] = 0` or

`(z')/ sqrt[1 - z] = 3 xx 8/9=8/3` or

`(dz)/ sqrt[1 - z] =8/3dx` or after integrating

`-2 sqrt(1-z) = 8/3 x + C` and

`(4/3x+C_1)^2=1-z rArr z = 1-(4/3x+C_1)^2 = (2/3 y)^2` and then

`y = pm3/2sqrt( 1-(4/3x+C_1)^2 )`

Now with the initial conditions

`y(0)=pm 3/2 sqrt(1-C_1^2) = 0 rArr C_1= pm1` and

`y = pm3/2sqrt( 1-(4/3xpm1)^2 )`