Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 16 maths

Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 16 maths

Answers (1)

Answer: \mathrm{c} y^{4}=e^{-x / y}

Given: The curve is y=f(x). SupposeP(x, y) is a part of the curve.

To find: We have to find the equation of the curve for which the intercept cut off by a tangent on x-axis.

Hint: Use equation of tangent i.e.Y-y=\frac{d y}{d x}(X-x) then solve the differential equation.

Solution: Equation of tangent to the curve at P is

=Y-y=\frac{d y}{d x}(X-x)

Where (X, Y) is the arbitrary point on the tangent

Putting Y=0 we get

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

When cut off by the tangent on the x-axis

=x-y \frac{d x}{d y}=4 y

Therefore   -y \frac{d x}{d y}=4 y-x

        \begin{aligned} &=\frac{d x}{d y}=\frac{x-4 y}{y} \\\\ &=\frac{d y}{d x}=\frac{y}{x-4 y} \ldots(i) \end{aligned}

This is homogeneous differential equation

Putting y=v x and \frac{d y}{d x}=v+x \frac{d v}{d x}  in (i)

We get

        \begin{aligned} &=v+x \frac{d v}{d x}=\frac{v x}{x-4 v x}=\frac{v x}{x(1-4 v)} \\\\ &=v+x \frac{d v}{d x}=\frac{v}{1-4 v} \\\\ &=x \frac{d v}{d x}=\frac{v}{1-4 v}-v \end{aligned}

        \begin{aligned} &=x \frac{d v}{d x}=\frac{v-v(1-4 v)}{1-4 v} \\\\ &=x \frac{d v}{d x}=\frac{v-v+4 v^{2}}{1-4 v} \\\\ &=x \frac{d v}{d x}=\frac{4 v^{2}}{1-4 v} \end{aligned}

        \begin{aligned} &=\frac{1}{x} \frac{d x}{d v}=\frac{1-4 v}{4 v^{2}} \\\\ &=4 \frac{d x}{x}=\frac{1-4 v}{v^{2}} d v \end{aligned}

Integrating on both sides,

        \begin{aligned} &=4 \int \frac{d x}{x}=\int \frac{1-4 v}{v^{2}} d v \\\\ &=4 \int \frac{d x}{x}=\int \frac{1}{v^{2}} d v-\int \frac{4 v}{v^{2}} d v \\\\ &=4 \int \frac{d x}{x}=\int \frac{1}{v^{2}} d v-4 \int \frac{d v}{v^{2}} \end{aligned}

        \begin{aligned} &=4 \log x+\log C=-\frac{1}{v}-4 \log v \\\\ &=4 \log x+4 \log v+\log C=-\frac{1}{v} \\\\ &=4 \log (x v)+\log C=-\frac{1}{v} \end{aligned}

Substituting value of v we get

        \begin{aligned} &=4 \log \left(x \times \frac{y}{x}\right)+\log C=-\frac{x}{y} \\\\ &=4 \log y+\log C=-\frac{x}{y} \end{aligned}

        =\log y^{4}+\log C=-\frac{x}{y} \quad\quad\quad\quad\left[a \log x=\log x^{a}\right]

        \begin{aligned} &=\log \left(y^{4} C\right)=-\frac{x}{y} \\\\ &=y^{4} C=e^{-\frac{x}{y}} \end{aligned}

Hence, required curve is found.

Posted by

infoexpert26

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question