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Question Number 45352 by Meritguide1234 last updated on 12/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

$$x-y=\frac{x^2}{y^2}$$$$y(\frac{x}{y}-1)=\frac{x^2}{y^2}t=\frac{x}{y}$$$$y(t-1)=t^2$$$$\frac{x}{t}(t-1)=t^2$$$$x=\frac{t^3}{t-1}y=\frac{x}{t}=\frac{t^2}{t-1}$$$$dx=\frac{(t-1)\left(3t^2\right)-t^3\left(1\right)}{{(t-1)}^2}dt$$$$dx=\frac{3t^3-3t^2-t^3}{{(t-1)}^2}dt$$$$\int \frac{dx}{3y-2x}$$$$=\int \frac{(2t^3-3t^2)}{{(t-1)}^2(\frac{3t^2}{t-1}-\frac{2t^3}{t-1})}dt$$$$\int \frac{(t-1)(2t^3-3t^2)}{{(t-1)}^2\times -(2t^3-3t^2)}dt$$$$=-1\int \frac{dt}{t-1}$$$$=-ln(t-1)+c$$$$=-ln(\frac{x}{y}-1)+c$$$$=-ln(x-y)+lny+c$$$$=ln\left(\frac{y}{x-y}\right)+cplscheck...$$$$$$$$$$

Commented by Meritguide1234 last updated on 12/Oct/18

$$yescorrect...myr.h.sshouldbe(x-y)indenominator.$$

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