Answer to Question #170016 in Differential Equations for Hiralal

Question

yp+x^2q^2=2x^2y

Answer 1

$yp + x^2q^2 = 2x^2y\\ f(x,y,z,p,q) = yp + x^2q^2 - 2x^2y = 0 \cdot \cdot \cdot (*) \\ f_x = 2xq^2 - 4xy\\ f_y = p- 2x^2\\ f_z = 0 \\ f_p = y \\ f_q = 2x^2q$

$\dfrac{dx}{-f_p}= \dfrac{dy}{-f_q} = \dfrac{dz}{-pf_p-qf_q} = \dfrac{dp}{f_x + pf_z} = \dfrac{dq}{f_y + qf_z}$

$\dfrac{dx}{-y}= \dfrac{dy}{-2x^2q} = \dfrac{dz}{-yp-2x^2q^2} = \dfrac{dp}{2xq^2-4xy} = \dfrac{dq}{p-2x^2}$

$\dfrac{dx}{-y}= \dfrac{dy}{-2x^2q} \\ ydy = 2x^2qdx \\ q= \frac{3}{4}x^{-3}(y^2 + c)$

Substituting into (*), we have

$yp + x^2 \bigg[\frac{3}{4}x^{-3}(y^2 + c) \bigg]^2 - 2x^2y = 0 \\ yp = 2x^2y - \frac{9}{16}x^{-4}(y^2 +c)^2 \\ p = 2x^2 - \frac{9}{16}x^{-4}y^{-1}(y^2 +c)^2$

Substituting the value of p and q into dz = pdx +qdy , we have

$dz = \bigg[2x^2 - \frac{9}{16}x^{-4}y^{-1}(y^2 +c)^2 \bigg] dx + \bigg[\frac{3}{4}x^{-3}(y^2 + c) \bigg]dy$

$z = \frac{2}{3}x^3+ \frac{3}{16}x^{-3}y^{-1}(y^2 +c)^2 + \frac{3}{4}x^{-3}(\frac{1}{3}y^3 + cy)+ d$

where c and d are arbitrary constant