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  • 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.

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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

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