Answer to Question #151078 in Differential Equations for fortune

Question

1.(x-y+1)dx+(x+4y+1)dy=0
2.(6x-3y+2)dx+(2x-y-1)dy=0

Answer 1

1) The given equation can be written as $\frac{dy}{dx}=-\frac{(x-y+1)}{(x+4y+1)}$ ......(1)

Here the two lines $x-y+1=0$ and $x+4y+1=0$ intersect at the point $(-1,0)$ .

We now transfer the origin by putting $x=-1+X,$ $y=0+Y$ ......(2)

This gives $dx=dX$ and $dy=dY$

Therefore from (1) , we get $\frac{dY}{dX}=-\frac{(X-Y)}{(X+4Y)}$ ........(3)

Put $Y=vX.$ Then the equation (3) becomes

$v+X.\frac{dv}{dX}=-\frac{(1-v)}{(1+4v)}$

$\implies X.\frac{dv}{dX}=-\frac{(1-v)}{(1+4v)}-v$

$\implies X.\frac{dv}{dX}=-\frac{(1+4v^2)}{(1+4v)}$

Separating the variables, we get

$\frac{(1+4v)}{(1+4v^2)}dv+\frac{dX}{X}=0$

Now integrating we get,

$\intop\frac{(1+4v)}{(1+4v^2)}dv+\intop \frac{dX}{X}=0$

$\implies \intop\frac{1}{(1+4v^2)}dv+\intop\frac{4v}{(1+4v^2)}dv+\intop \frac{dX}{X}=0$

$\implies \frac {1}{2}.tan^{-1}(2v)+\frac{1}{2}.log|4v^2+1|+log|X|=log|c|$ $[$where $log|c|$ is an integrating constant$]$

$\implies tan^{-1}(2v)+log|4v^2+1|+2log|X|=2log|c|$

$\implies \tan^{-1}(2.\frac{Y}{X})+log|\frac{4Y^2+X^2}{X^2}|+log|X|^2=log|c|^2$ [Putting the value of $v=\frac{Y}{X}$]

$\implies \tan^{-1}(2.\frac{Y}{X})+log|{4Y^2+X^2}|-log|{X^2}|+log|X|^2=log|c|^2$

$\implies \tan^{-1}(2.\frac{Y}{X})+log|{4Y^2+X^2}|=K(say)$

$\implies \tan^{-1}(\frac{2y}{x+1})+log|{4y^2+(x+1)^2}|=K$ [ Putting the value $X=x+1$ and $Y=y$ ]

Which is the required solution.

2) Here the two lines $6x-3y+2=0$ and $2x-y-1=0$ are parallel.

We see that the given equation may be written as $\frac{dy}{dx}=-[\frac{3(2x-y)+2}{(2x-y)-1}].$ ........(1)

Put $2x-y=v$ in (1) , so that $\frac{dv}{dx}=2-\frac{dy}{dx}$

Hence we obtain from (1), $2-\frac{dv}{dx}=-[\frac{3v+2}{v-1}]$

$\implies \frac{dv}{dx}=[\frac{3v+2}{v-1}]+2$

$\implies \frac{dv}{dx}=[\frac{5v}{v-1}]$

Separating the variables we get , $\frac{v-1}{5v}dv=dx$ .......(2)

Now integrating (2) we have , $\intop \frac{v-1}{5v}dv=\intop dx$

$\implies \frac{1}{5}\intop dv- \frac{1}{5}\intop \frac{1}{v}dv=\intop dx$

$\implies \frac{1}{5}v-\frac{1}{5}log|v|=x+C$ [where $C$ is an integrating constant]

$\implies v-log|v|=5x+5C$

$\implies (2x-y)-log|2x-y|=5x+5C$ [ Putting $v=2x-y$ ]

$\implies log|2x-y|+(3x+y)=K$ (say)

Which is the required solution.