Solve the following differential equations: dy dx =y 2x, y(0)=1 - Vidyadip

Question

Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x, y}(0)=1$

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Answer

We have,
$\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x, y}(0)=1$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\sin2\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\sin2\text{x dx}$
$\Rightarrow\log|\text{y}|=-\frac{\cos 2\text{x}}{2}+\text{C}...(1)$
Given: $\text{x}=0,\text{y}=1.$
Substituting the values of x and y in (1), we get
$\log|1|=-\frac{1}{2}+\text{C}$
$\Rightarrow\text{C}=\frac{1}{2}$
Substituting the values of C in (1), we get
$\log|\text{y}|=-\frac{\cos 2\text{x}}{2}+\frac{1}{2}$
$\Rightarrow\log|\text{y}|=\frac{1-\cos 2\text{x}}{2}$
$\Rightarrow\log|\text{y}|=\sin^2\text{x}$
$\Rightarrow\text{y}=\text{e}^{\sin^2\text{x}}$
Hence, $\text{y}=\text{e}^{\sin^2\text{x}}$ is the required solution

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