Question

Find the initial value problem in `dy/dx=(y^2+4x^2)/(4xy)`?

Addition information `y(1)=2`

Answer 1

`y= sqrt(4/3x^2+8/3sqrtx)`

Explanation:

Given: `dy/dx=(y^2+4x^2)/(4xy), y(1)=2`

Rewrite in the form of Bernoulli's equation:

`dy/dx - 1/(4x)y= x/y`

Prepare for a variable substitution by multiplying both sides by `2y`:

`2ydy/dx - 1/(2x)y^2= 2x`

Let `u = y^2`, then `(du)/dx= 2ydy/dx`:

`(du)/dx - 1/(2x)u= 2x`

This type of differential equation is known to have an integrating factor of:

`I(x) = e^(int -1/(2x)dx)`

`I(x) = e^(-1/2ln(x))`

`I(x) = e^(ln(1/sqrtx))`

`I(x) = 1/sqrtx`

Multiply both sides by the integrating factor:

`(du)/dx1/sqrtx - 1/2x^(-3/2)u= 2sqrtx`

We know that the left side integrates to `I(x)u` and the right side is trivial:

`u/sqrtx= 4/3x^(3/2)+C`

`u= 4/3x^2+Csqrtx`

Reverse the substitution `u = y^2`:

`y^2= 4/3x^2+Csqrtx`

`y= +-sqrt(4/3x^2+Csqrtx)`

Use the boundary condition `y(1) = 2` to determine `+-` and the value of C:

`2= +-sqrt(4/3 1^2+Csqrt1)`

It must be `+`:

`2= sqrt(4/3 1^2+Csqrt1)`

`4 = 4/3+C`

`4 = 4/3+C`

`C = 8/3`

`y= sqrt(4/3x^2+8/3sqrtx)`