Question
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}\text{y}^2,\text{y}(0)=-1$
✓
Answer
$\frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}\text{y}^2,\text{y}(0)=-1$
$\Rightarrow\frac{1}{\text{y}^2}\text{dy}=2\text{e}^{2\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^2}\text{dy}=2\int\text{e}^{2\text{x}}\text{dx}$
$\Rightarrow\frac{-1}{\text{y}}=\text{e}^{2\text{x}}+\text{C}...(1)$
We know that at $\text{x}=0,\text{y}=-1.$
Substituting the values of x and y in (1), we get
$1=1+\text{C}$
$\Rightarrow\text{C}=0$
Substituting the value of C in (1), we get
$-\frac{1}{\text{y}}=\text{e}^{2\text{x}}$
$\Rightarrow\text{y}=-\text{e}^{-2\text{x}}$
Hence, $\text{y}=-\text{e}^{-2\text{x}}$ is the required soluton.
$\Rightarrow\frac{1}{\text{y}^2}\text{dy}=2\text{e}^{2\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^2}\text{dy}=2\int\text{e}^{2\text{x}}\text{dx}$
$\Rightarrow\frac{-1}{\text{y}}=\text{e}^{2\text{x}}+\text{C}...(1)$
We know that at $\text{x}=0,\text{y}=-1.$
Substituting the values of x and y in (1), we get
$1=1+\text{C}$
$\Rightarrow\text{C}=0$
Substituting the value of C in (1), we get
$-\frac{1}{\text{y}}=\text{e}^{2\text{x}}$
$\Rightarrow\text{y}=-\text{e}^{-2\text{x}}$
Hence, $\text{y}=-\text{e}^{-2\text{x}}$ is the required soluton.
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