- $\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
- $\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{C}$
- $\text{e}^{-\text{x}}+\text{e}^{\text{y}}=\text{C}$
- $\text{e}^{-\text{x}}+\text{e}^{-\text{y}}=\text{C}$
$\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
Solution:
We have,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}\times\text{e}^{\text{y}}$
$\Rightarrow \text{e}^{-\text{y}}\text{dy}=\text{e}^{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\text{e}^{-\text{y}}\text{dy}=\int\text{e}^{\text{x}}\text{dx}$
$\Rightarrow \text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{D}$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^{-\text{y}}=-\text{D}$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^{-\text{y}}=-\text{C}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
(i) $x \times y = b$.....
(ii) $x\,.\,a = 1$.....
(iii)Then $x = .........,\,\,\,y = .......$