The curve represented by the differential equation $\frac{d y}{d x} = \frac{2(x+1) y}{x^2+2 x+2}, y>0$, passes through the point $(0,4)$. If it also passes through $P(-1, k)$ and $Q(\lambda, 10)$ then $(P Q)^2$ is equal to

68

$\frac{d y}{d x}=\frac{2(x+1) y}{x^2+2 x+2} \Rightarrow \frac{1}{y} d y=\frac{2(x+1) d x}{x^2+2 x+1+1^2}$

$\Rightarrow \int \frac{1}{y} d y=\int \frac{2(x+1)}{(x+1)^2+1} d x$

Let Z = $(x + 1)^2 + 1$

$dz = 2(x + 1) dx$

$\int \frac{1}{y} d y=\int \frac{d z}{z}$

$\Rightarrow \log y=\log z+\log c$

$\Rightarrow \log y=\log z c \Rightarrow y=zc$

$\Rightarrow y = ((x+1)^2 + 1) c$    as it passes (0, 4)

4 = (1² + 1)c

$\Rightarrow c=\frac{4}{2}=2$         so y = [(x + 1)² + 1] × 2

at P(-1, k)  ⇒  k = $((-1+1)^2 + 1) × 2) ⇒  k = 2$

so P(-1, 2)

Q(λ, 10)  ⇒  10 = $((λ+1)^2 + 1) × 2) ⇒  5 = (λ+1)^2 + 1$

⇒  $4 = (λ+1)^2$

so  λ + 1 = ±2

λ = 1, -3

Q(1, 10)   or   Q(-3, 10)

$(PQ)^2 = \left(\sqrt{(1+1)^2+(10-2)^2}\right)^2$  or    $\left(\sqrt{(-3+1)^2+(10-2)^2}\right)^2$

$\Rightarrow(P Q)^2=68$