Question

The particular solution of $\frac{y}{x}\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$, when $x=1,y=2$ is:

• A. $5(1+y^2)=2(1+x^2)$
• B. $2(1+y^2)=5(1+x^2)$
• C. $2(1+y^2)=(1+x^2)$
• D. $(1+y^2)=2(1+x^2)$

Answer 1

Answer: (B)

$2(1+y^2)=5(1+x^2)$

We have $\frac{ydy}{xdx}=\frac{1+y^2}{1+x^2}$

$\frac{ydy}{1+y^2}=\frac{xdx}{1+x^2}$

Multiplying both sides with 2
$\frac{2ydy}{1+y^2}=\frac{2xdx}{1+x^2}$

$\int \frac{2ydy}{1+y^2}=\int \frac{2xdx}{1+x^2}$

$ln(1+y^2)=ln(1+x^2)+ln\left(C\right)$

$(1+y^2)=C(1+x^2)$
Now,

$y=2$ at $x=1$. Hence $2C=5$.
Therefore

$2(1+y^2)=2C(1+x^2)$

$2(1+y^2)=5(1+x^2)$