Q50 / 155 · WBJEE 2019 · Mathematics

General solution of $(x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0$ is $(C$ is an arbitrary constant)

  • A
    $\frac{x}{a}=\tan \frac{y}{a}+C$
  • B
    $\tan x y=C$
  • C
    $\tan (x+y)=C$
  • Correct answer
    D
    $\tan \frac{y+C}{a}=\frac{x+y}{a}$

Solution

$(x+y)^{2} \frac{d y}{d x}=a^{2}, a \neq 0$ Let $x+y=t$ $1+\frac{d y}{d x}=\frac{d t}{d x}$ $\Rightarrow \quad \frac{d y}{d x}=\frac{d t}{d x}-1$ Therefore, $t^{2}\left(\frac{d t}{d x}-1\right)=a^{2}$ $\Rightarrow \quad t^{2} \frac{d t}{d x}-t^{2}=a^{2}$ $\Rightarrow \quad t^{2} \frac{d t}{d x}=a^{2}+t^{2}$ $\Rightarrow \quad \frac{t^{2}}{a^{2}+t^{2}} d t=d x$ [variable separation $]$ $\Rightarrow \quad \int \frac{t^{2}}{\left(t^{2}+a^{2}\right)} d t=\int d x$ [integration] $\Rightarrow \quad \int \frac{t^{2}+a^{2}-a^{2}}{\left(t^{2}+a^{2}\right)} d t=x+C'$ [doing $a^{2}$ adding and subtracting] $\Rightarrow \quad \int\left(1-\frac{a^{2}}{t^{2}+a^{2}}\right) d t=x+C'$ $\Rightarrow \quad \int d t-\int \frac{a^{2}}{t^{2}+a^{2}} d t=x+C'$ $\Rightarrow \quad t-a^{2} \int \frac{d t}{t^{2}+a^{2}}=x+C^{\prime}$ $\Rightarrow t-a^{2}\left[\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)\right]=x+C'$ $\Rightarrow \quad(x+y)-a \tan ^{-1}\left(\frac{x+y}{a}\right)$ $=x+C'$ [on putting the value of t] $\Rightarrow \quad \frac{y-C^{\prime}}{a}=\tan ^{-1}\left(\frac{x+y}{a}\right)$ $\Rightarrow \quad \frac{x+y}{a}=\tan \left(\frac{y-C^{\prime}}{a}\right)$ $\Rightarrow \frac{x+y}{a}=\tan \left(\frac{y+c}{a}\right)$ [where, $C=-C^{\prime}$ is an arbitrary constant $]$ $\Rightarrow \quad \tan \left(\frac{y+C}{a}\right)=\frac{x+y}{a}$

Hence, the correct answer is option: 'D'.

Relevant Questions