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Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is `y = 2(e^(x) - x-1)`. |
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Answer» We have , `(dy)/(dx) = y+2x` `rArr (dy)/(dx) - y = 2x` `IF = e^(-int dx) = e^(-x)` Multiplying the equation by IF and integrating `y xx e^(-x) = int 2xe^(-x) dx + k`, k being the constant of integration. `= 2[x int e^(x) dx - int 1.(-e^(x))dx] + k`, using integration by parts `= 2-xe^(-x) - 2e^(x) + k` ...(i) As curve (i) passes through (0, 0) 0 = 0 - 2 + k `:. k = 2` Thus the curve is `ye^(-x) = -2xe^(-x) - 2e^(-x) + 2` `:. y = -2x - 2 + 2e^(x)` `:. y = 2(e^(x) - x-1)` |
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