For the differential equation


For the differential equation $x y \frac{d y}{d x}=(x+2)(y+2)$, find the solution curve passing through the point $(1,-1)$.


The differential equation of the given curve is:

$x y \frac{d y}{d x}=(x+2)(y+2)$

$\Rightarrow\left(\frac{y}{y+2}\right) d y=\left(\frac{x+2}{x}\right) d x$


$\Rightarrow\left(1-\frac{2}{y+2}\right) d y=\left(1+\frac{2}{x}\right) d x$

Integrating both sides, we get:

$\int\left(1-\frac{2}{y+2}\right) d y=\int\left(1+\frac{2}{x}\right) d x$

$\Rightarrow \int d y-2 \int \frac{1}{y+2} d y=\int d x+2 \int \frac{1}{x} d x$

$\Rightarrow y-2 \log (y+2)=x+2 \log x+\mathrm{C}$

$\Rightarrow y-x-\mathrm{C}=\log x^{2}+\log (y+2)^{2}$

$\Rightarrow y-x-\mathrm{C}=\log \left[x^{2}(y+2)^{2}\right]$                 ...(1)

Now, the curve passes through point (1, –1).

$\Rightarrow-1-1-C=\log \left[(1)^{2}(-1+2)^{2}\right]$

$\Rightarrow-2-C=\log 1=0$

$\Rightarrow C=-2$

Substituting $C=-2$ in equation $(1)$, we get:

$y-x+2=\log \left[x^{2}(y+2)^{2}\right]$

This is the required solution of the given curve.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now