Question

The solution of $\left(x+y+1\right)\frac{dy}{dx}=1$ is

• A. $y=\left(x+2\right)+c{e}^x$
• B. $y=-\left(x+2\right)+c{e}^x$
• C. $x=-\left(y+2\right)+c{e}^y$
• D. $x={\left(y+2\right)}^2+c{e}^y$

Answer 1

The correct option is C

$x=-\left(y+2\right)+c{e}^y$

Explanation of the correct option:

The given differential equation: $\left(x+y+1\right)\frac{dy}{dx}=1$.

Let us assume that, $x+y+1=z$

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(x+y+1\right)=\frac{dz}{dx} \\ \Rightarrow \frac{dx}{dx}+\frac{dy}{dx}+0=\frac{dz}{dx} \\ \Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1 \end{gathered}$$

Consider the equation: $\left(x+y+1\right)\frac{dy}{dx}=1$

$$\begin{gathered} \Rightarrow z\left(\frac{dz}{dx}-1\right)=1 \\ \Rightarrow \frac{dz}{dx}-1=\frac{1}{z} \\ \Rightarrow \frac{dz}{dx}=\frac{1}{z}+1 \\ \Rightarrow \frac{dz}{dx}=\frac{1+z}{z} \\ \Rightarrow dx=\frac{zdz}{z+1} \\ \Rightarrow dx=\frac{\left(z+1-1\right)dz}{z+1} \\ \Rightarrow dx=dz-\frac{dz}{z+1} \end{gathered}$$

Integrate both sides of the equation.

$$\begin{gathered} \Rightarrow \int dx=\int dz-\int \frac{dz}{z+1} \\ \Rightarrow x=z-\log \left|z+1\right|+c_1 \\ \Rightarrow x=x+y+1-\log \left|x+y+2\right|+c_1\mathbf{[}\mathbf{\because }\mathbf{z}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{1}\mathbf{]} \\ \Rightarrow -y=-\log \left|x+y+2\right|+c_1+1 \\ \Rightarrow y=\log \left|x+y+2\right|-\left(c_1+1\right) \\ \Rightarrow y+\left(c_1+1\right)=\log \left|x+y+2\right| \\ \Rightarrow e^{y+\left(c_1+1\right)}=x+y+2 \\ \Rightarrow e^{c_1+1}{e}^y=x+y+2 \\ \Rightarrow x=-\left(y+2\right)+c{e}^y\mathbf{}\left[\because c=e^{c_1+1}\right] \end{gathered}$$

Therefore, the solution of the differential equation $\left(x+y+1\right)\frac{dy}{dx}=1$ is $x=-\left(y+2\right)+c{e}^y$.

Hence, option C is correct.