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Differential Equations

Question
CBSEENMA12033214

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.

Solution
Let y = f (x) be the equation of curve.
We know that dy over dx represents the slope of the tangent to the curve at the point (x, y).
From the given condition.
                            dy over dx space equals space straight x plus straight y              or          dy over dx minus straight y space equals straight x
Comparing dy over dx plus straight P space straight y space equals space straight Q space space space with space dy over dx minus straight y space equals space straight x comma space we space get comma space space straight P space equals space minus 1 comma space space straight Q space equals space straight x
therefore space space space space space space space space integral straight P space dx space equals space minus integral 1 space dx space equals space minus straight x comma space space space straight e to the power of integral straight P space dx end exponent space equals space straight e to the power of negative straight x end exponent
Solution of differential equation is
                       straight y space straight e to the power of integral straight P space dx end exponent space equals integral straight Q space straight e to the power of integral straight P space dx end exponent dx space plus space straight c
or              straight y space straight e to the power of negative straight x end exponent space equals space integral space straight x space straight e to the power of negative straight x end exponent space dx plus straight c space space space or space space space space straight y space straight e to the power of negative straight x end exponent space equals space straight x fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction minus integral 1. space fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction dx plus straight c
or                straight y space straight e to the power of negative straight x end exponent space equals space minus straight x space straight e to the power of negative straight x end exponent plus integral space straight e to the power of negative straight x end exponent dx plus straight c
or                straight y space straight e to the power of negative straight x end exponent space equals space minus xe to the power of negative straight x end exponent plus straight e to the power of negative straight x end exponent plus straight c
or                 straight y space equals space minus straight x minus 1 plus straight c over straight e to the power of negative straight x end exponent
or            straight x plus straight y plus 1 space equals space straight c space straight e to the power of straight x                                  ...(1)
Now curve passes through origin (0,  0).
therefore space space space 0 plus 0 plus 1 space equals space straight c space straight e to the power of 0 space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight c space equals space 1
therefore     from (1),  straight x plus straight y plus 1 space equals space straight e to the power of straight x is the required equation of curve. 

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