Math, asked by Uvwuwve, 1 month ago

(2ydx+3xdy) +2xy(3ydx+4xdy) =0

Answers

Answered by hukam0685
3

Step-by-step explanation:

Given:

(2ydx + 3xdy) + 2xy(3ydx + 4xdy) = 0 \\

To find: Solve and find dy/dx

Solution:

Step 1: Open bracket

2ydx + 3xdy + 6x {y}^{2} dx + 8 {x}^{2} ydy = 0 \\

Step 2: Rearrange the terms

3xdy + 8 {x}^{2} ydy + 2ydx + 6x {y}^{2} dx = 0 \\

Step 3: Take dy common from first two terms and dx from last two

dy(3x + 8 {x}^{2} y) + dx(2y + 6x{y}^{2} ) = 0 \\

Step 4: Take dx and it's coefficient to RHS

dy(3x + 8 {x}^{2} y)  =  -  dx(2y + 6x{y}^{2} )  \\

Step 5: Cross multiply

 \frac{dy}{dx}  = -   \frac{2y + 6x  {y}^{2} }{3x + 8 {x}^{2}y }  \\  \\

Final answer:

 \bold{ \red{\frac{dy}{dx}  = -   \frac{2y + 6x  {y}^{2} }{3x + 8 {x}^{2}y } }} \\  \\

Hope it helps you.

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