Question

(2ydx + 3xdy) + 2xy(3ydx + 4xdy) = 0​

Answer 1

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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