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