Solving and checking a first order Bernoulli differential equation: $\dfrac{\mathrm{d}y}{\mathrm{d}x} + y = y^2 $

Question

Problem:

Solve the following differential equation: $$ \dfrac{dy}{dx} + y = y^2 $$

Answer:

Let $v = y^{-1}$. \begin{align*} \dfrac{dv}{dx} &= -y^{-2} \dfrac{dy}{dx} \\ -y^{-2}\dfrac{dy}{dx} - y^{-1}&= -1 \\ \dfrac{dv}{dx} - v &= -1 \\ \end{align*} Now we need to find an integrating factor. Call the integrating factor $I$. \begin{align*} I &= e^{ \int -dx } = e^{-x} \\ e^{-x} \dfrac{dv}{dx} - e^{-x}v &= -1 \\ e^{-x}v &= -x + C \\ v &= (c-x)e^x \\ y^{-1} &= (c-x)e^x \\ \end{align*} Hence the answer is: $$ y(c-x)e^x = 1 $$ Now, I wil check my answer. \begin{align*} y &= \dfrac{1}{(c-x)e^x} = \left( (c-x)e^x \right) ^{-1}\\ \dfrac{dy}{dx} &= (-1)\left( (c-x)e^x - e^x \right) ^{-2} \\ \dfrac{dy}{dx} &= (-1)\left( (c -1 - x)e^x \right) ^{-2} \\ \dfrac{dy}{dx} &= \dfrac{ e^{2x} }{ (c-1-x)^2 } \\ \dfrac{dy}{dx} + y &= \dfrac{ e^{2x} }{ (c-1-x)^2 } + \left( (c-x)e^x \right) ^{-1} \end{align*} I do not see how I can show that: $$ \dfrac{dy}{dx} + y = y^2 $$ Hence my solution does not check. Where did I go wrong?

Here is an updated soltion which I believe is right. Is it?

Let $v = y^{-1}$. \begin{align*} \dfrac{dv}{dx} &= -y^{-2} \dfrac{dy}{dx} \\ -y^{-2}\dfrac{dy}{dx} - y^{-1}&= -1 \\ \dfrac{dv}{dx} - v &= -1 \\ \end{align*} Now we need to find an integrating factor. Call the integrating factor $I$. \begin{align*} I &= e^{ \int -dx } = e^{-x} \\ e^{-x} \dfrac{dv}{dx} - e^{-x}v &= -e^{-x} \\ e^{-x}v &= e^{-x} + C \\ v &= 1 + Ce^{x} \\ y^{-1} &= 1 + Ce^{x} \\ \end{align*} Hence the answer is: $$ y = (1 + Ce^{x})^{-1} $$ Now, I wil check my answer. \begin{align*} \dfrac{dy}{dx} &= (-1)(Ce^x)\left( 1 + Ce^{x} \right) ^{-2} \\ \dfrac{dy}{dx} &= -Ce^x \left( 1 + Ce^{x} \right) ^{-2} \\ \dfrac{dy}{dx} + y &= -Ce^x \left( 1 + Ce^{x} \right) ^{-2} + (1 + Ce^{x})^{-1} \\ % \dfrac{dy}{dx} + y &= \dfrac{ -Ce^x + 1 + Ce^x}{ \left( 1 + Ce^{x} \right) ^ 2 } \\ % \dfrac{dy}{dx} + y &= \dfrac{ 1}{ \left( 1 + Ce^{x} \right) ^ 2 } \\ \dfrac{dy}{dx} + y &= y^2 \end{align*} Hence my solution checks.

Answer 1

After finding the integrating factor, you should have multiplied the entire equation by $I = e^{-x}$. Instead you only multiplied the left. The correct equation is $$e^{-x} \frac{\mathrm dv}{\mathrm dx} - e^{-x} v = e^{-x}.$$

Answer 2

HINT

Alternatively, you can notice the proposed ODE is separable: \begin{align*} y' + y = y^{2} & \Longleftrightarrow y' = y^{2} - y \Longleftrightarrow \frac{y'}{y^{2} - y} = 1 \Longleftrightarrow \frac{y'}{\left(y - \frac{1}{2}\right)^{2} - \frac{1}{4}} = 1 \end{align*}

Can you take it from here?