Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 72 textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 72 textbook solution.

Answers (1)

Answer :x^{2}-y^{2}=c x

Given : \frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}

Hint:

Using variable separable method.

Explanation:

We know that the slope of the tangent at(x,y) of a Curve is  \frac{dy}{dx}

 

Given slope of tangent at (x,y) is \frac{x^{2}+y^{2}}{2 x y}

\therefore \frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}-(1)

\begin{aligned} &\text { Put } \frac{d y}{d x}=f(x, y) \Rightarrow F(x, y)=\frac{x^{2}+y^{2}}{2 x y} \\ &\text { Finding } f(\lambda x, \lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{2(\lambda x)(\lambda y)} \\ \end{aligned}

                                        \begin{aligned} &=\frac{\lambda^{2} x^{2}+\lambda^{2} y^{2}}{2 \lambda^{2} x y} \end{aligned}

                                       \begin{aligned} &=\frac{\lambda^{2}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)}{\lambda^{2} \cdot 2 \mathrm{xy}} \\ \end{aligned}

                                      \begin{aligned} &=\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{2 \mathrm{xy}} \\ \end{aligned}

                                      \begin{aligned} &=\mathrm{F}(\mathrm{x}, \mathrm{y}) \\ \end{aligned}

\begin{aligned} &\text { So, } \mathrm{F}(\lambda \mathrm{x}, \lambda \mathrm{y})=\mathrm{F}(\mathrm{x}, \mathrm{y}) \\ \end{aligned}

                              \begin{aligned} &=\lambda^{\circ} \mathrm{F}(\mathrm{x}, \mathrm{y}) \end{aligned}

So \mathrm{F}(\lambda \mathrm{x}, \lambda \mathrm{y}) is homogeneous function of degree 0 

\therefore Given equation is homogeneous differential equation

Now Put y =vx in (1) 

Differentiate with respect to x

\begin{aligned} &\frac{d y}{d x}=x \frac{d v}{d x}+v\\ &\text { Putting } \frac{\mathrm{dy}}{\mathrm{dx}} \text { value in }(1) \text { and } \mathrm{y}=\mathrm{vx} \end{aligned}

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

\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\frac{1-v^{2}}{2 v} \\ &\Rightarrow \frac{d v}{d x}=\frac{1-v^{2}}{2 v} \cdot \frac{1}{x} \Rightarrow \frac{2 v}{1-v^{2}} d v=\frac{d x}{x} \end{aligned}

\begin{aligned} &\Rightarrow-\frac{2}{v^{2}-1} d v=\frac{d x}{x} \\ &\Rightarrow \frac{2 v d v}{v^{2}-1}=-\frac{d x}{x} \end{aligned}

Integrating both sides,

\begin{aligned} &\int \frac{2 \mathrm{v}}{\mathrm{v}^{2}-1} \mathrm{dv}=-\int \frac{\mathrm{dx}}{\mathrm{x}} \\ &\Rightarrow \int \frac{\partial \mathrm{v}}{\mathrm{v}^{2}-1} \mathrm{~d} \mathrm{v}=-\int \frac{\mathrm{dx}}{\mathrm{x}} \\ &\Rightarrow \int \frac{\partial \mathrm{v}}{\mathrm{v}^{2}-1} \mathrm{dv}=-\log |\mathrm{x}|+\mathrm{c} \end{aligned}

Solving \int \frac{2v}{v^{2}-1}dv

Put v^{2}-1=t

Differentiate with respect to v

\begin{gathered} 2 v d v=d t \\ d v=\frac{d t}{2 v} \\ =\int \frac{2 v}{t} \cdot \frac{d t}{2 v} \\ =\int \frac{d t}{t}=\log 1+1 \end{gathered}

Putting back

\begin{aligned} &\mathrm{t}=\mathrm{v}^{2}-1 \\ &\Rightarrow \log \left|\mathrm{v}^{2}-1\right| \end{aligned}

\therefore By (2)

\begin{aligned} &\log \left|\mathrm{v}^{2}-1\right|=-\log |\mathrm{x}|+\mathrm{c}_{1}\\ &\text { Putting }\\ &\mathrm{vx}=\mathrm{y} \text { or } \end{aligned}

\begin{aligned} &\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}} \\ &\log \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-1\right|=-\log |\mathrm{x}|+\mathrm{C}_{1} \\ &\Rightarrow \log \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-1\right|+\log |\mathrm{x}| \\ &=\mathrm{C}_{1} \end{aligned}

\begin{aligned} &\Rightarrow \log \left|\left[\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-1\right] \dot{\mathrm{x}}\right| \\ &=\mathrm{C}_{1} \\ &\Rightarrow \log \left|\left[\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}-1\right] \mathrm{x}\right|=\mathrm{C}_{1} \end{aligned}

Putting C_{1}=\log\; C_{1}

\log \left|\left(\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}-1\right) \mathrm{x}\right|=\log \mathrm{C}_{1}

Remove log of both sides.

\begin{aligned} &\\\left(\frac{y^{2}}{x^{2}}-1\right) x=C_{1} \\ \end{aligned}

\begin{aligned} \\\Rightarrow & \frac{x y^{2}}{x^{2}}-x=C_{1} \\ \end{aligned}

\begin{aligned} \\\Rightarrow \frac{y^{2}}{x}-x=C_{1} \end{aligned}

\begin{aligned} &\Rightarrow \mathrm{y}^{2}-\mathrm{x}^{2}=\mathrm{x} \mathrm{C}_{1} \\ \\&\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}=-\mathrm{C}_{1} \mathrm{x} \\ \\&\text { Put } \mathrm{C}=-\mathrm{C}_{1} \\ \\&\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{cx} \end{aligned}

Hence Proved

 

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question