question_answer

Find the equation of a curve passing through the point (0, 1), if the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate (abscissa) and the product of the x-coordinate and y-coordinate (ordinate) of that point.

Answer:

We know that slope of tangent to the curve y = f(x)at any point (x, y) is \[\frac{dy}{dx}.\] According to the given condition, we have             \[\frac{dy}{dx}=x+xy\] or \[\frac{dy}{dx}-xy=x\] Which is a linear differential equation of the from             \[\frac{dy}{dx}+Py=Q.\] Where, \[P=-\,x\,\,\text{and}\,\,Q+x\] Now, \[IF={{e}^{\int{Pdx}}}={{e}^{\int{(-x)dx}}}={{e}^{-{{x}^{2}}/2}}\] Now, the solution is given by \[y\cdot (IF)=\int{(Q\cdot IF)dx+C}\] \[\Rightarrow \]   \[y\cdot {{e}^{-{{x}^{2}}/2}}=\int{x\cdot {{e}^{-{{x}^{2}}/2}}dx+C}\] Now, put \[\frac{{{x}^{2}}}{2}=t\] \[\Rightarrow \] \[xdx=dt\] \[\therefore \]      \[y\cdot {{e}^{-{{x}^{2}}/2}}=\int{{{e}^{-t}}dt+C}\] \[\Rightarrow \]   \[y\cdot {{e}^{-{{x}^{2}}/2}}=-\,{{e}^{-t}}+C\] \[\Rightarrow \]   \[y\cdot {{e}^{-{{x}^{2}}/2}}=-\,{{e}^{-{{x}^{2}}/2}}+C\] \[\Rightarrow \]   \[y=-\,1+C{{e}^{{{x}^{2}}/2}}\]                             ?(i) Since, the curve passes through the point (0, 1). \[\therefore \] We have, \[1=-\,1+Ce{}^\circ \] \[\Rightarrow \] C = 2 Hence, the required equation of curve is             \[y=-\,1+2{{e}^{{{x}^{2}}/2}}\]