Question

(2xy+1)y DX + (1+2xy-x³y³)x dy = 0​

Answer 1

SOLUTION

TO SOLVE

$\displaystyle\sf{(2xy + 1)y \: dx + (1 + 2xy - {x}^{3} {y}^{3})x \: dy = 0 }$

EVALUATION

Here the given differential equation is

$\displaystyle\sf{(2xy + 1)y \: dx + (1 + 2xy - {x}^{3} {y}^{3}) x\: dy = 0 }$

$\displaystyle\sf{ \implies \: 2x {y}^{2}dx + y \: dx + xdy+ 2 {x}^{2} ydy - {x}^{4} {y}^{3}\: dy = 0 }$

$\displaystyle\sf{ \implies \: 2x {y}^{2}dx + 2 {x}^{2} ydy+ y \: dx + xdy - {x}^{4} {y}^{3}\: dy = 0 }$

$\displaystyle\sf{ \implies \: 2xy( ydx + xdy)+ (y \: dx + xdy )- {x}^{4} {y}^{3}\: dy = 0 }$

$\displaystyle\sf{ \implies \: 2xy \: d(xy) + d(xy) - {x}^{4} {y}^{3}\: dy = 0 }$

$\displaystyle\sf{ \implies \: \frac{2xy \: d(xy)}{ {x}^{4} {y}^{4} } + \frac{d(xy)}{ {x}^{4} {y}^{4} } - \frac{{x}^{4} {y}^{3}\: dy}{ {x}^{4} {y}^{4} } = 0 }$

$\displaystyle\sf{ \implies \: \frac{2 \: d(xy)}{ {x}^{3} {y}^{3} } + \frac{d(xy)}{ {x}^{4} {y}^{4} } - \frac{dy}{ y} = 0 }$

Let z = xy

Then from above we get

$\displaystyle\sf{ \frac{2 \: dz}{ {z}^{3} } + \frac{dz}{ {z}^{4} } - \frac{dy}{ y} = 0 }$

$\displaystyle\sf{ \implies \: 2 {z}^{ - 3}dz+ {z}^{ - 4}dz - \frac{dy}{ y} = 0 }$

On integration we get

$\displaystyle\sf{ \int \: 2 {z}^{ - 3}dz+ \int{z}^{ - 4}dz - \int \frac{dy}{ y} = 0 }$

$\displaystyle\sf{ \implies \: - \frac{ 1}{ {z}^{2} } - \frac{1}{3 {z}^{3} } - \log \: |y| = - c }$

$\displaystyle\sf{ \implies \: \frac{ 1}{ {x}^{2} {y}^{2} } + \frac{1}{3 {x}^{3} {y}^{3} } + \log \: |y| = c }$

Where C is integration constant

FINAL ANSWER

Hence the required solution is

$\boxed{ \: \: \displaystyle\sf{ \: \frac{ 1}{ {x}^{2} {y}^{2} } + \frac{1}{3 {x}^{3} {y}^{3} } + \log \: |y| = c } \: \: }$

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Answer 2

Step-by-step explanation:

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