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  • Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Q17.    Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

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Let f(x , y)  is the curve passing through point (0 , 2)
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} +5= y + x \\ \\ \frac{dy}{dx}-y=x-5
It is   \frac{dy}{dx}+py=Q   type of equation where p = -1 \ and \ Q = x- 5
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-5) \times e^{-x} )dx+ C
Now, Let
I= \int ((x-5) \times e^{-x} )dx \\ \\ I = (x-5).\int e^{-x}dx-\int \left ( \frac{d(x-5)}{dx}.\int e^{-x}dx \right )dx\\ \\ I = -(x-5)e^{-x}+\int e^{-x}dx\\ \\ I = -xe^{-x}-e^{-x}+5e^{-x}\\ \\ I = -e^{-x}(x-4)
Put this value in our equation 
ye^{-x}= -e^{-x}(x-4)+C
Now, by using boundary conditions we will find the value of C
It is given that curve passing through   point  (0 , 2)
2.e^{-0}= -e^{-0}(0-4)+C\\ \\ C = -2
Our final equation becomes
ye^{-x}= -e^{-x}(x-4)-2\\ y=4-x-2e^x
Therefore, the required equation of curve is y=4-x-2e^x

Posted by

Gautam harsolia

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