1. ## Find particular solution

Find the particular solution of the differential equation

$\displaystyle \frac{y}{x}\frac{dy}{dx}=\frac{y^2+1}{x^2+1}$ y=0 when x=1

I solved the differential equation by first separating the variables:

$\displaystyle \int \frac{y}{y^2+1}dy=\int \frac{x}{x^2+1}dx$

$\displaystyle \frac{1}{2}\ln |y^2+1|=\frac{1}{2}\ln |x^2+1|+A$ where A=the arbitrary constant.

Now i don't know how to proceed.
Thanks

2. Originally Posted by arze
Find the particular solution of the differential equation
$\displaystyle \frac{y}{x}\frac{dy}{dx}=\frac{y^2+1}{x^2+1}$ y=0 when x=1
I solved the differential equation by first separating the variables
$\displaystyle \int \frac{y}{y^2+1}dy=\int \frac{x}{x^2+1}dx$
$\displaystyle \frac{1}{2}\ln |y^2+1|=\frac{1}{2}\ln |x^2+1|+A$ where A=the arbitrary constant.
Now i don't know how to proceed.
Thanks
Absolute value bars aren't necessary since both $\displaystyle x^2+1>0$ and $\displaystyle y^2+1>0\,\forall x,y\in\mathbb{R}$.

From there, multiply both sides by 2 to get $\displaystyle \ln\left(y^2+1\right)=\ln\left(x^2+1\right)+A$ (note that 2A is a new constant [why not rename it A?])

Now, apply the exponential functions to both sides to get $\displaystyle y^2+1=e^{\ln\left(x^2+1\right)+A}=Ce^{\ln\left(x^2 +1\right)}=C\left(x^2+1\right)$.

Now solve for $\displaystyle y^2$ to get $\displaystyle y^2=C\left(x^2+1\right)-1$

Its best to leave this in implicit form to avoid the plus/minus issue.