Question
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\text{y}\tan2\text{x, y}(0)=2$
✓
Answer
We have,
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan2\text{x, y}(0)=2$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\tan2\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\tan2\text{x dx}$
$\Rightarrow\log|\text{y}|=\frac{1}{2}\log|\sec2\text{x}|+\frac{1}{2}\log\text{C}$
$\Rightarrow\text{y}^2=\text{C}\sec2\text{x}\dots(1)$
It is given that at $\text{x}=0,\text{y}=2.$
$\therefore\text{C}=4$
Substituting the value of C in (1), we get
$\therefore\text{y}^2=\frac{4}{\cos2\text{x}}$
$\Rightarrow\text{y}=\frac{2}{\sqrt{\cos2\text{x}}}$
Hence, $\text{y}=\frac{2}{\sqrt{\cos2\text{x}}}$ is the required solution.
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan2\text{x, y}(0)=2$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\tan2\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\tan2\text{x dx}$
$\Rightarrow\log|\text{y}|=\frac{1}{2}\log|\sec2\text{x}|+\frac{1}{2}\log\text{C}$
$\Rightarrow\text{y}^2=\text{C}\sec2\text{x}\dots(1)$
It is given that at $\text{x}=0,\text{y}=2.$
$\therefore\text{C}=4$
Substituting the value of C in (1), we get
$\therefore\text{y}^2=\frac{4}{\cos2\text{x}}$
$\Rightarrow\text{y}=\frac{2}{\sqrt{\cos2\text{x}}}$
Hence, $\text{y}=\frac{2}{\sqrt{\cos2\text{x}}}$ is the required solution.
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