Question

(x^3y^2+x)dy + (x^2y^3-y)dx =0​

Answer 1

SOLUTION

TO SOLVE

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

EVALUATION

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

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

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

$\displaystyle \sf{ \implies \: {x}^{2} {y}^{2} d(xy )= ydx - xdy}$

Dividing both sides by xy we get

$\displaystyle \sf{ \implies \: xy \: d(xy )= \frac{dx}{x} - \frac{dy}{y} }$

Let z = xy

$\displaystyle \sf{ \implies \: z \: dz= \frac{dx}{x} - \frac{dy}{y} }$

On integration we get

$\displaystyle \sf{ \implies \int z \: dz= \int\frac{dx}{x} - \int\frac{dy}{y} }$

$\displaystyle \sf{ \implies \frac{ {z}^{2} }{2} = log(x) - log(y) + c}$

$\displaystyle \sf{ \implies \frac{ {x}^{2} {y}^{2} }{2} = log(x) - log(y) + c}$

Where C is integration constant

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