Question

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 (x + 1) = 2e^2x.

Answer 1

According to the question, 
\[\frac{dy}{dx} = y + 2x\]
\[ \Rightarrow \frac{dy}{dx} - y = 2x . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
where P = - 1 and Q = 2x
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{- \int dx} \]
\[ = e^{- x} \]
\[\text{Multiplying both sides of }\left( 1 \right)\text{ by }I . F . = e^{- x} , \text{ we get }\]
\[ e^{- x} \left( \frac{dy}{dx} - y \right) = e^{- x} 2x \]
\[ \Rightarrow e^{- x} \frac{dy}{dx} - e^{- x} y = e^{- x} 2x \]
Integrating both sides with respect to x, we get

\[ \Rightarrow y e^{- x} = 2x\int e^{- x} dx - 2\int\left[ \frac{d}{dx}\left( x \right)\int e^{- x} dx \right]dx + C\]
\[ \Rightarrow y e^{- x} = - 2x e^{- x} - 2 e^{- x} + C . . . . . \left( 2 \right)\]
Since the curve passes through origin, we have
\[0 \times e^0 = - 2 \times 0 \times e^0 - 2 e^0 + C\]
\[ \Rightarrow C = 2\]
\[\text{ Putting the value of C in }\left( 2 \right),\text{ we get }\]
\[y e^{- x} = - 2x e^{- x} - 2 e^{- x} + 2\]
\[ \Rightarrow y = - 2x - 2 + 2 e^x \]
\[ \Rightarrow y + 2\left( x + 1 \right) = 2 e^x \]