Answer to Question #161089 in Differential Equations for Sri

Question

solve y(1+2xycosx^2-2xy)dx+(sinx^2-x^2)dy=0

Answer 1

NOTE: I asked for cosx^2 is cos(x²) or cos²x at the work start but no answer.

Solution

y(1+2xycos²x-2xy)dx + (sin²x-x²)dy = 0

Dividing by (sin²x-x²)dx we will get

dy/dx = -y(1+2xycos²x-2xy)/(sin²x-x²) =

           = y(2xysin²x - 1)/(sin²x-x²) = -y/(sin²x-x²)+ 2x y² sin²x /(sin²x-x²)

So y’+ y/(sin²x-x²) = [2x sin²x /(sin²x-x²)] y²

It is Bernoulli differential equation

y’ + p(x)y = q(x)y²  

with p(x) = (sin²x-x²)^-1 , q(x) = 2x sin²x /(sin²x-x²) 

Divide this by y² .

y^-2y’ + p(x)y^-1 = q(x)   

The substitution and derivative that we’ll need here is

v=y⁻¹, v′=−y⁻²y′

With this substitution the differential equation becomes

v’ – p(x)v = -q(x)

Let m(x) = exp(-∫p(x) dx) = exp(-∫(sin²x-x²)^-1  dx)

Multiplying last equation m(x) we’ll get

(m(x)v)’ = -m(x) q(x)

Integrating this ∫ (m(x)v)’dx = -∫m(x) q(x)dx and so

v = 1/y = -[m(x)]^-1∫ m(x) q(x)dx - C

And y = -{[m(x)]^-1∫ m(x) q(x)dx + C}^-1  

Finally

y = -{[ exp(-∫(sin²x-x²)^-1  dx)]^-1∫ exp(-∫(sin²x-x²)^-1  dx)*2x sin²x /(sin²x-x²) dx + C}^-1

or

$y(x)=\ -\{ \ [ exp(-\int \frac{dx}{sin^2x-x^2} ) \ ]^{-1} \int exp(-\int \frac{dx}{sin^2x-x^2} ) \frac{2x sin^2x}{sin^2x-x^2} dx +C \}^{-1}$