Find the equation of the curve passing through (1, 1) and the slope of the tangent to curve at a point (x, y) is equal to the twice the sum of the abscissa and the ordinate.

Explanation

`because` The slope of the tangent at (x, y) is `(dy)/(dx)`
`:. (dy)/(dx) = 2(x+y)` ...(1)
Let x + y = t
`rArr 1 + (dy)/(dx) = (dt)/(dx)`
`:.` From equation (1)
`(dt)/(dx) - 1 = 2t`
`rArr (dt)/(2t + 1) = dx`
Integrating both sides, we get
`(1)/(2) ln (2t + 1) = x + ln c_(1)`
`rArr ln(2t + 1) = 2x + ln c`, where `ln c = 2ln c_(1)`
`rArr 2t + 1 = ce^(2x)`
or `2(x + y) + 1 = ce^(2x)`
`because` The curve passes through (1, 1)
`:. 2(1+1) + 1= ce^(2)`
`rArr c = 5e^(-2)`
`:.` The equation of the curve is `2(x+y) + 1 = 5e^(2(x-1))`