Solving the Differential Equation

The problem asks for the solution to the differential equation $\frac{dy}{dx} = y \sin 2x$ given the initial condition $y(0) = 1$. This is a first-order separable differential equation.

Separating the Variables

First, we rearrange the equation to separate the terms involving $y$ and the terms involving $x$. We can divide both sides by $y$ (assuming $y \neq 0$) and multiply by $dx$:

$ \frac{1}{y} dy = \sin 2x \, dx $

Integrating Both Sides

Next, we integrate both sides of the equation:

$ \int \frac{1}{y} dy = \int \sin 2x \, dx $

The integral of $\frac{1}{y}$ with respect to $y$ is $\ln|y|$.

The integral of $\sin 2x$ with respect to $x$ is $-\frac{1}{2} \cos 2x$.

So, the integration yields:

$ \ln|y| = -\frac{1}{2} \cos 2x + C_1 $

where $C_1$ is the constant of integration.

Solving for y

To find $y$, we exponentiate both sides using the base $e$:

$ e^{\ln|y|} = e^{-\frac{1}{2} \cos 2x + C_1} $ $ |y| = e^{C_1} e^{-\frac{1}{2} \cos 2x} $

Let $C = \pm e^{C_1}$. Since $e^{C_1}$ is always positive, $C$ can be any non-zero constant. We can also include the case $y=0$ by allowing $C=0$. Thus, the general solution is:

$ y = C e^{-\frac{1}{2} \cos 2x} $

Applying the Initial Condition

We are given the initial condition $y(0) = 1$. We substitute $x=0$ and $y=1$ into the general solution to find the value of $C$:

$ 1 = C e^{-\frac{1}{2} \cos (2 \times 0)} $ $ 1 = C e^{-\frac{1}{2} \cos 0} $

Since $\cos 0 = 1$, we have:

$ 1 = C e^{-\frac{1}{2} (1)} $ $ 1 = C e^{-1/2} $

Solving for $C$:

$ C = \frac{1}{e^{-1/2}} = e^{1/2} $

Finding the Particular Solution

Now, we substitute the value of $C$ back into the general solution:

$ y = e^{1/2} \cdot e^{-\frac{1}{2} \cos 2x} $

Using the property $e^a \cdot e^b = e^{a+b}$:

$ y = e^{\frac{1}{2} - \frac{1}{2} \cos 2x} $ $ y = e^{\frac{1}{2} (1 - \cos 2x)} $

Simplifying Using Trigonometric Identities

We can simplify the exponent using the double-angle identity $1 - \cos 2x = 2 \sin^2 x$:

$ y = e^{\frac{1}{2} (2 \sin^2 x)} $ $ y = e^{\sin^2 x} $

Conclusion

The required solution to the differential equation $\frac{dy}{dx} = y \sin 2x$ with the initial condition $y(0) = 1$ is $y = e^{\sin^2 x}$.