Question

the integrating factor of x2ydx-(x3+y3)dy=0 is​

Answer 1

Step-by-step explanation:

We have,

$\tt \: {x}^{2} ydx - ( {x}^{3} + {y}^{3} )dy = 0$

$\tt \: \implies \frac{dx}{dy} = \frac{ {x}^{3} + {y}^{3}}{ {x}^{2}y }$

$\tt \: \implies \frac{dx}{dy} = \frac{x}{y} + \frac{ {y}^{2} }{ {x}^{2} }$

$\tt \: \implies \frac{dx}{dy} - \frac{x}{y} = \frac{ {y}^{2} }{ {x}^{2} }$

$\tt \: \implies {x}^{2} \frac{dx}{dy} - \frac{ {x}^{3} }{y} = {y}^{2}$

Put $\sf\:x^3=v$

$\sf\implies\:3x^2\frac{dx}{dy}=\frac{dv}{dy}$

So,

$\tt \: \implies \frac{1}{3} \frac{dv}{dy} - \frac{ v}{y} = {y}^{2}$

$\tt \: \implies \frac{dv}{dy} - \frac{3 }{y} v = 3{y}^{2}$

$\sf \: I.F. = {e}^{ \int - \frac{3}{y}dy } = {e}^{ - 3ln(y)} = \frac{1}{ {y}^{3} }$

So,

$\tt \frac{1}{ {y}^{3} } .v = \int \frac{1}{ {y}^{3} } \cdot \: 3 {y}^{2} dy$

$\tt \implies\frac{v}{ {y}^{3} } = \int \frac{3}{y } dy$

$\tt \implies\frac{v}{ {y}^{3} } = 3 ln(y) + C$

$\tt \implies\frac{ {x}^{3} }{ {y}^{3} } = ln(y^{3} ) + C$