The solution of (x - y^3)dx + 3xy^2dy = 0 is | vedclass

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(B) Given differential equation is $(x - y^3)dx + 3xy^2dy = 0$.Rearranging the terms,we get $x dx - y^3 dx + 3xy^2 dy = 0$.Let $t = y^3$,then $dt = 3y

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$\log x + \frac{y^3}{x} = k$

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(B) Given differential equation is $(x - y^3)dx + 3xy^2dy = 0$.Rearranging the terms,we get $x dx - y^3 dx + 3xy^2 dy = 0$.Let $t = y^3$,then $dt = 3y^2 dy$.Substituting these into the equation,we get $x dx - t dx + x dt = 0$.Dividing by $x^2$,we get $\frac{x dx - t dx + x dt}{x^2} = 0$,which simplifies to $\frac{dx}{x} + \frac{x dt - t dx}{x^2} = 0$.This is equivalent to $d(\log x) + d(\frac{t}{x}) = 0$.Integrating both sides,we get $\log x + \frac{t}{x} = k$.Substituting $t = y^3$ back,the solution is $\log x + \frac{y^3}{x} = k$.

The solution of $(x - y^3)dx + 3xy^2dy = 0$ is

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