Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Q16.    Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Answers (1)

best_answer

Let f(x , y)  is the curve passing through origin 
Then, the slope of tangent to the curve at point (x , y) is given by  \frac{dy}{dx}
Now, it is given that
\frac{dy}{dx} = y + x\\ \\ \frac{dy}{dx}-y=x
It is   \frac{dy}{dx}+py=Q   type of equation where p = -1 \ and \ Q = x
Now,
I.F. = e^{\int pdx}= e^{\int -1dx }= e^{-x}
Now,
y(I.F.)= \int (Q \times I.F. )dx+ C
y(e^{-x})= \int (x \times e^{-x} )dx+ C
Now, Let
I= \int (x \times e^{-x} )dx \\ \\ I = x.\int e^{-x}dx-\int \left ( \frac{d(x)}{dx}.\int e^{-x}dx \right )dx\\ \\ I = -xe^{-x}+\int e^{-x}dx\\ \\ I = -xe^{-x}-e^{-x}\\ \\ I = -e^{-x}(x+1)
Put this value in our equation 
ye^{-x}= -e^{-x}(x+1)+C
Now, by using boundary conditions we will find the value of C
It is given that curve passing through origin i.e. (x , y) = (0 , 0)
0.e^{-0}= -e^{-0}(0+1)+C\\ \\ C = 1
Our final equation becomes
ye^{-x}= -e^{-x}(x+1)+1\\ y+x+1=e^x
Therefore, the required equation of the curve is y+x+1=e^x

Posted by

Gautam harsolia

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

CBSE announces STEM education as annual training theme for 2025
anam-khan

Board Exam 2025 Admit Card LIVE: CBSE Class 10, 12 hall ticket out; when will state boards issue admit cards?
anam-khan

CBSE Class 10, 12 admit card 2025 out for regular, private students at Pariksha Sangam portal

Latest Question