The equation of the curve which passes through point $(1,0)$ and has tangent with slope…
The equation of the curve which passes through point $(1,0)$ and has tangent with
slope $1+\frac{y}{x}+\left(\frac{y}{x}\right)^{2}$ is
- $\tan ^{-1}\left(\frac{x}{y}\right)=\log |x|$
- $\tan ^{-1}\left(\frac{x}{y}\right)=\log |y|$
- $\tan ^{-1}\left(\frac{y}{x}\right)=\log |y|$
- $\tan ^{-1}\left(\frac{y}{x}\right)=\log |x|$
Solution
We have siope of tangent $=\frac{d y}{d x}=1+\frac{y}{x}+\left(\frac{y}{x}\right)^{2}$
$\therefore \frac{d y}{d x}=1+\frac{y}{x}+\frac{y^{2}}{x^{2}} \Rightarrow \frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}}$ ...(1)
Put $y=u x \Rightarrow \frac{d y}{d x}=u+x \frac{d u}{d x}$
$\therefore \quad u+x \frac{d u}{d x}=\frac{x^{2}+u x^{2}+u^{2} x^{2}}{x^{2}}$
$\therefore \quad u+x \frac{d u}{d x}=1+u+u^{2} \Rightarrow x \frac{d u}{d x}=1+u^{2} \Rightarrow \int \frac{d u}{1+u^{2}}=\int \frac{d x}{x}$
$\therefore \tan ^{-1} u=\log |x|+c \Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\log |x|+c$
At $(1,0)$, we write $\tan ^{-1}(0)=\log |1|+c \Rightarrow c=0$
$\therefore \tan ^{-1}\left(\frac{y}{x}\right)=\log |x|$
Asked in: MHT CET 2020 (13 Oct Shift 1)
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