#### Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 67 Subquestion (ii) textbook solution.

Answer : $\tan ^{-1} \frac{y}{x}+\log \left|\sqrt{\frac{y^{2}}{x^{2}}}+1 \times x\right|=\frac{\pi}{4}+\frac{1}{2} \log 2$

Hint                 : using variable separable method and substituting the values

Given              :   $(x+y) d y+(x-y) d x=0 \quad, y=1 \text { when } x=1$

Solution :

\begin{aligned} &(x+y) d y+(x-y) d x=0 \\ &\Rightarrow(x+y) d y=-(x-y) d x \\ &\Rightarrow(x+y) d y=(y-x) d x \\ &\Rightarrow \frac{d y}{d x}=\frac{y-x}{x+y}=\frac{-(x-y)}{x+y} \end{aligned}

Now let ,

\begin{aligned} &f(x, y)=\frac{d y}{d x}=\frac{-(x-y)}{x+y} \\ &\text { Finding } f(\lambda x, \lambda y) \end{aligned}

\begin{aligned} f(\lambda x, \lambda y) &=\frac{-(\lambda x-\lambda y)}{\lambda x+\lambda y} \\ &=\frac{-\lambda(x-y)}{\lambda(x+y)} \\ &=\frac{-(x-y)}{x+y} \\ &=\lambda^{0} f(x, y) \end{aligned}

Therefore, $f(x,y)$ is a homogeneous function of degree 0 .

Putting $y=vx$

Diff w.r.t.x

\begin{aligned} &\frac{d y}{d x}=x \frac{d v}{d x}+v\\ &\text { Putting value of } \frac{d y}{d x} \text { and } y=v x \text { in (i) } \end{aligned}

$\begin{gathered} \frac{d y}{d x}=\frac{-(x-y)}{x+y} \\ v+x \frac{d v}{d x}=\frac{-(x-v x)}{x+v x} \\ v+x \frac{d v}{d x}=\frac{-x(1-v)}{x(1+v)} \end{gathered}$

$\begin{gathered} =\frac{v-1}{1+v} \\ \Rightarrow x \frac{d v}{d x}=\frac{v-1}{1+v}-v \\ \Rightarrow x \frac{d v}{d x}=\frac{v-1-v(1+v)}{1+v} \end{gathered}$

\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\frac{-\left(1+v^{2}\right)}{1+v} \\ &\Rightarrow \frac{1+v}{1+v^{2}} d v=-\frac{d x}{x} \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{1+v}{1+v^{2}} d v=-\int \frac{d x}{x} \\ &\Rightarrow \int \frac{1}{1+v^{2}} d v+\int \frac{v}{v^{2}+1} d v=-\log |x|+c \\ &\Rightarrow \tan ^{-1} v+\int \frac{v}{v^{2}+1} d v=-\log |x|+c \end{aligned}

Putting $v^{2}+1=t$ in integral

\begin{aligned} &2 v d v=d t \\ &v d v=\frac{d t}{2} \end{aligned}

\begin{aligned} &\Rightarrow \tan ^{-1} v+\int \frac{1}{t} \times \frac{d t}{2}=-\log |x|+c \\ &\Rightarrow \tan ^{-1} v+\frac{1}{2} \log |t|=-\log |x|+c \\ &\Rightarrow \tan ^{-1} v+\frac{1}{2} \log \left|v^{2}+1\right|=-\log |x|+c \end{aligned}

Now again putting back the value of  $v=\frac{y}{x}$

\begin{aligned} &\tan ^{-1} \frac{y}{x}+\frac{1}{2} \log \left|\left(\frac{y}{x}\right)^{2}+1\right|+\log |x|=c \\ &\Rightarrow \tan ^{-1} \frac{y}{x}+\log \left|\sqrt{\left(\frac{y}{x}\right)^{2}+1}\right|+\log |x|=c \\ &\Rightarrow \tan ^{-1} \frac{y}{x}+\log \left|\sqrt{\left(\frac{y}{x}\right)^{2}+1} \times x\right|=c \end{aligned}                                            ....(ii)

Now, $y = 1$ when $x = 1$

\begin{aligned} &\text { Therefore } \tan ^{-1}\left(\frac{1}{1}\right)+\log \left|\sqrt{\left(\frac{1}{1}\right)^{2}+1} \times 1\right|=c \\ &\tan ^{-1}(1)+\log \sqrt{2}=c \\ &\frac{\pi}{4}+\frac{1}{2} \log 2=c \end{aligned}

Put in (ii)

$\tan ^{-1} \frac{y}{x}+\log \left|\sqrt{\frac{y^{2}}{x^{2}}+1} \times x\right|=\frac{\pi}{4}+\frac{1}{2} \log 2$