- $\text{e}^{\text{y}}=\text{e}^{\text{x}^{2}}+\text{C}$
Solution:
We have
$\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}-\text{y}}$ $\Rightarrow\text{e}^{\text{y}}=\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}}$
$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=2\int\text{x}\text{e}^{\text{x}^{2}}\text{dx}$
Put $\text{x}^2=\text{t}$ in R.H.S. integral we get
$2\text{x}\text{dx}=\text{dt}$
$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=\int\text{e}^{\text{t}}\text{dt}$
$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{t}}+\text{C}$
$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{x}^2}+\text{C}$