Question
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=2\text{e}^{\text{x}}\text{y}^3,\text{y}(0)=\frac{1}{2}$
$\frac{\text{dy}}{\text{dx}}=2\text{e}^{\text{x}}\text{y}^3,\text{y}(0)=\frac{1}{2}$
✓
Answer
We have,
$\frac{\text{dy}}{\text{dx}}=2\text{e}^{\text{x}}\text{y}^3,\text{y}(0)=\frac{1}{2}$
$\Rightarrow\frac{1}{\text{y}^3}\text{dy}=2\text{e}^{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^3}\text{dy}=\int2\text{e}^{\text{x}}\text{dx}$
$\Rightarrow-\frac{1}{2\text{y}^3}=2\text{e}^{\text{x}}+\text{C}...(1)$
Given: at $\text{x}=0,\text{y}=\frac{1}{2}$
Substituting the valuse of x and y in (1), we get
$-\frac{1}{2\times\frac{1}{4}}=2\text{e}^{0}+\text{C}$
$\Rightarrow\text{C}=-2-2$
$\Rightarrow\text{C}=-4$
Substituting the value of C in (1), we get
$\Rightarrow-\frac{1}{2\text{y}^2}=2\text{e}^{\text{x}}-4$
$\Rightarrow\text{y}^{2}(8-4\text{e}^{\text{x}})=1$
Hence, $\text{y}^{\text{x}}(8-4\text{e}^{\text{x}})=1$ is the required solution.
$\frac{\text{dy}}{\text{dx}}=2\text{e}^{\text{x}}\text{y}^3,\text{y}(0)=\frac{1}{2}$
$\Rightarrow\frac{1}{\text{y}^3}\text{dy}=2\text{e}^{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}^3}\text{dy}=\int2\text{e}^{\text{x}}\text{dx}$
$\Rightarrow-\frac{1}{2\text{y}^3}=2\text{e}^{\text{x}}+\text{C}...(1)$
Given: at $\text{x}=0,\text{y}=\frac{1}{2}$
Substituting the valuse of x and y in (1), we get
$-\frac{1}{2\times\frac{1}{4}}=2\text{e}^{0}+\text{C}$
$\Rightarrow\text{C}=-2-2$
$\Rightarrow\text{C}=-4$
Substituting the value of C in (1), we get
$\Rightarrow-\frac{1}{2\text{y}^2}=2\text{e}^{\text{x}}-4$
$\Rightarrow\text{y}^{2}(8-4\text{e}^{\text{x}})=1$
Hence, $\text{y}^{\text{x}}(8-4\text{e}^{\text{x}})=1$ is the required solution.
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