MCQ
Choose the correct answer from the given four options.The general solution of $\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}-\text{y}}$ is:
  • A
    $\text{e}^{\text{x}^{2}-\text{y}}=\text{C}$
  • B
    $\text{e}^{-\text{y}}+\text{e}^{\text{x}^{2}}=\text{C}$
  • $\text{e}^{\text{y}}=\text{e}^{\text{x}^{2}}+\text{C}$
  • D
    $\text{e}^{\text{x}^{2}+\text{y}}=\text{C}$

Answer

Correct option: C.
$\text{e}^{\text{y}}=\text{e}^{\text{x}^{2}}+\text{C}$
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

$2\text{x}\text{dx}=\text{dt}$

$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=\int\text{e}^{\text{t}}\text{dt}$

$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{t}}+\text{C}$

$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{x}^2}+\text{C}$

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