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

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

Answers (1)

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

 

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