Question

If $\mathrm{y}=\mathrm{y}\left(\mathrm{x}\right)$ is the solution of the differential equation ,${\mathrm{e}}^{\mathrm{y}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-1\right)={\mathrm{e}}^{\mathrm{x}}$such that $\mathrm{y}\left(0\right)=0$, then $\mathrm{y}\left(1\right)$ is equal to :

• A. ${\log }_{e}2$
• B. $2e$
• C. $2+{\log }_{e}2$
• D. $1+{\log }_{e}2$

Answer 1

The correct option is D

$1+{\log }_{e}2$

Explanation for the correct option:

Step 1: Simplifying differential equation:

The given differential equations are ${\mathrm{e}}^{\mathrm{y}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-1\right)={\mathrm{e}}^{\mathrm{x}}...\left(\mathrm{i}\right)$

$\mathrm{y}\left(0\right)=0...\left(\mathrm{ii}\right)$

From $\left(\mathrm{i}\right),$differentiating on both sides,

$\begin{array}{c}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-1\right)=\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{y}}}\\ ={\mathrm{e}}^{\mathrm{x}-\mathrm{y}}...\left(\mathrm{iii}\right)\end{array}$

Assuming $\mathrm{x}-\mathrm{y}=\mathrm{z}$

Differentiating on both sides, with respect to $\mathrm{x}$

$$\begin{gathered} 1-\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dz}}{\mathrm{dx}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}-1=-\frac{\mathrm{dz}}{\mathrm{dx}}...\left(\mathrm{iv}\right) \end{gathered}$$

From $\left(\mathrm{iii}\right)$ and $\left(\mathrm{iv}\right)$,we get

$\begin{array}{l}\frac{-\mathrm{dz}}{\mathrm{dx}}={\mathrm{e}}^{\mathrm{z}}\\ \Rightarrow -\frac{1}{{\mathrm{e}}^{\mathrm{z}}}\mathrm{dz}=\mathrm{dx}\\ \Rightarrow -{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}=\mathrm{dx}...\left(\mathrm{v}\right)\end{array}$

Integrating the equation $\left(\mathrm{v}\right)$

$$\begin{gathered} {\mathrm{e}}^{-\mathrm{z}}=\mathrm{x}+\mathrm{c} \\ {\mathrm{e}}^{-(\mathrm{x}-\mathrm{y})}=\mathrm{x}+\mathrm{c}\left[\because z=x-y\right] \end{gathered}$$

Step 2: Find the value of $c$:

Substituting $\mathrm{x}=0$ and $\mathrm{y}\left(0\right)=0$

$$\begin{gathered} {\mathrm{e}}^{-(0-\mathrm{y}(0\left)\right)}=0+\mathrm{c} \\ \Rightarrow {\mathrm{e}}^{-(0-0)}=0+\mathrm{c} \\ \Rightarrow {\mathrm{e}}^0=\mathrm{c} \\ \Rightarrow \mathrm{c}=1 \end{gathered}$$

$\therefore {\mathrm{e}}^{-(\mathrm{x}-\mathrm{y})}=\mathrm{x}+1$

Step 3: Finding the value of $\mathrm{y}\left(1\right)$.

Taking $\log$ on both sides,

$\begin{array}{l}\log \left({\mathrm{e}}^{-(\mathrm{x}-\mathrm{y})}\right)=\log \left(\mathrm{x}+1\right)\\ \Rightarrow -\left(\mathrm{x}-\mathrm{y}\right)=\log \left(\mathrm{x}+1\right)\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{loge}\mathbf{=}\mathbf{1}\mathbf{]}\\ \Rightarrow -\mathrm{x}+\mathrm{y}=\log \left(\mathrm{x}+1\right)\\ \Rightarrow \mathrm{y}=\mathrm{x}+\log \left(\mathrm{x}+1\right)\end{array}$

Place it $\mathrm{x}=1$now,

$\mathrm{y}\left(1\right)=1+\log (1+1)$

$\mathrm{y}\left(1\right)=1+{\log }_{\mathrm{e}}2$

Hence, the correct option is (D).