Question

Solve (ylogx-2)ydx-xdy=0

Answer 1

This is the Bernoulli's equation solution...

Answer 2

Solve the given differential equation $(ylogx-2)ydx-xdy=0$

Explanation:

• given is the differential equation in non- linear form. we convert it to linear differential equation of the form    $\frac{dt}{dx}+Pt=Q$    here, P and Q are functions of 'x' and t is a function of 'y'.  
• Its solution is given by , $t(I.F)=\int\ {Q(I.F)} \, dx$
• here I.F stands for integrating factor is given as, $I.F=e^{\int{P} \, dx }$
• hence now we have,                                                                                       $(ylogx-2)ydx-xdy=0\\y^2logx(dx)-2y(dx)-x(dy)=0\\(y^2xdx)(\frac{logx}{x}-\frac{2}{x}y^{-1}-y^{-2}\frac{dy}{dx})=0\\\\y^{-2}\frac{dy}{dx}+\frac{2}{x}y^{-1}=\frac{logx}{x}$
• now equating $y^{-1}=t\ \ \ \ \ \ \ \ ->y^{-2}\frac{dy}{dx} =-\frac{dt}{dx}$  in above equation,                       $-\frac{dt}{dx}+\frac{2t}{x}=\frac{logx}{x} \\->\frac{dt}{dx}-\frac{2t}{x}=-\frac{logx}{x}\ \ \ \ \ \ \ -----> \frac{dt}{dx}+Pt=Q$  
• hence we have , $P=\frac{-2}{x} \ \ \ \ Q=-\frac{logx}{x}$  we get the I.F and solution as,            $I.F=e^{\int {P} \, dx }=e^{-2\int {\frac{1}{x} } \, dx } \\I.F=e^{-2logx}= e^{logx^{-2}}\\->I.F=x^{-2} \\Solution\ of\ the\ equation\ is,\ \ \ \ t(I.F)=\int\ {Q(I.F)} \, dx \\->t(x^{-2})=\int\ {-\frac{logx}{x^3} }\,dx\\\\->\frac{t}{x^2}= \frac{2logx+1}{4x^2} +C$
• now putting the value of 't' back we get the final solution of the differential equation as  $\frac{1}{yx^2}=\frac{2logx+1}{4x^2}+C$