Question
Find the equation of the 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
According to the question,
\[\frac{dy}{dx} = x + y\]
\[\Rightarrow \frac{dy}{dx} - y = x\]
\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]
\[P = - 1 \]
\[Q = x\]
Now,
\[I . F . = e^{- \int dx} = e^{- x} \]
So, the solution is given by
\[y \times I . F . = \int Q \times I . F . dx + C\]
\[ \Rightarrow y e^{- x} = x\int e^{- x} dx - \int\left[ \frac{d}{dx}\left( x \right)\int e^{- x} dx \right]dx + C\]
\[ \Rightarrow y e^{- x} = - x e^{- x} + \int e^{- x} dx + C\]
\[ \Rightarrow y e^{- x} = - x e^{- x} - e^{- x} + C\]
Since the curve passes throught the origin, it satisfies the equation of the curve.
\[ \Rightarrow 0 e^0 = - 0 e^0 - e^0 + C\]
\[C = 1\]
Putting the value of C in the equation of the curve, we get
\[y e^{- x} = - x e^{- x} - e^{- x} + 1\]
\[ \Rightarrow y e^{- x} + x e^{- x} + e^{- x} = 1\]
\[ \Rightarrow \left( y + x + 1 \right) e^{- x} = 1\]
\[ \Rightarrow \left( x + y + 1 \right) = e^x\]
