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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. |
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Answer» `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))` |
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