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i have to solve the following $1^{st}$ order differential equation

$(xy+1)dx+(2y-x)dy=0$

i am in the elementary differential class,and have not learned multivariate functions, the equation below is none exact,since $M_y=x\ne N_x=-1$ so i am looking for a substitution that can make it exact because the intergration factor has both $x$ and $y$ ...

$\large \frac{dy}{dx}=\frac{xy+1}{x-2y}$

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Look into integrating factors. –  Git Gud Oct 17 '13 at 23:08

2 Answers 2

Approach $1$:

$(xy+1)~dx+(2y-x)~dy=0$

$(xy+1)~dx=(x-2y)~dy$

$(x-2y)\dfrac{dy}{dx}=xy+1$

Let $u=\dfrac{x}{2}-y$ ,

Then $y=\dfrac{x}{2}-u$

$\dfrac{dy}{dx}=\dfrac{1}{2}-\dfrac{du}{dx}$

$\therefore2u\left(\dfrac{1}{2}-\dfrac{du}{dx}\right)=x\left(\dfrac{x}{2}-u\right)+1$

$u-2u\dfrac{du}{dx}=\dfrac{x^2}{2}-xu+1$

$2u\dfrac{du}{dx}=(x+1)u-\dfrac{x^2}{2}-1$

$u\dfrac{du}{dx}=\dfrac{(x+1)u}{2}-\dfrac{x^2+2}{4}$

This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $u=\dfrac{1}{v}$ ,

Then $\dfrac{du}{dx}=-\dfrac{1}{v^2}\dfrac{dv}{dx}$

$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dx}=\dfrac{x+1}{2v}-\dfrac{x^2+2}{4}$

$\dfrac{dv}{dx}=\dfrac{(x^2+2)v^3}{4}-\dfrac{(x+1)v^2}{2}$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2

Approach $2$:

$(xy+1)~dx+(2y-x)~dy=0$

$(xy+1)~dx=(x-2y)~dy$

$(yx+1)\dfrac{dx}{dy}=x-2y$

Let $u=x+\dfrac{1}{y}$ ,

Then $x=u-\dfrac{1}{y}$

$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$

$\therefore yu\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=u-\dfrac{1}{y}-2y$

$yu\dfrac{du}{dy}+\dfrac{u}{y}=u-\dfrac{2y^2+1}{y}$

$yu\dfrac{du}{dy}=\dfrac{(y-1)u}{y}-\dfrac{2y^2+1}{y}$

$u\dfrac{du}{dy}=\dfrac{(y-1)u}{y^2}-\dfrac{2y^2+1}{y^2}$

This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $u=\dfrac{1}{v}$ ,

Then $\dfrac{du}{dy}=-\dfrac{1}{v^2}\dfrac{dv}{dy}$

$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dy}=\dfrac{y-1}{y^2v}-\dfrac{2y^2+1}{y^2}$

$\dfrac{dv}{dy}=\dfrac{(2y^2+1)v^3}{y^2}-\dfrac{(y-1)v^2}{y^2}$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2

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setting $y=vx$ we get

$dy=vdx+xdv$ and substituting we get

$(vx^2+1)dx+(2y-1)(vdx+xdv)=0$

$(vx^2+2yv-vx+1)dx+(2yx-x^2)dv=0$

$(vx^2+2v^2x-vx+1)dx+x^2(2v-1)dv=0$

$(vx(x+2v-1)+1)dx+x^2(2v-1)dv=0$

this should be seperable but i cannot seperate it easily...

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