Solve the differential equation dy/dx = sqrt(1 - 4y^2) with the initial condition y(0) = 0.75.

Steps to Solve

1. Separate the variables The given differential equation is $\frac{dy}{dx} = \sqrt{1 - 4y^2}$. To separate the variables, we can rewrite the equation as $\frac{dy}{\sqrt{1 - 4y^2}} = dx$.

2. Integrate both sides Now, we integrate both sides of the separated equation: $$ \int \frac{dy}{\sqrt{1 - 4y^2}} = \int dx $$

3. Solve the left-hand side integral To solve the integral on the left-hand side, we can use a substitution. Let $u = 2y$, so $du = 2dy$, and $dy = \frac{1}{2}du$. Then the integral becomes: $$ \int \frac{\frac{1}{2}du}{\sqrt{1 - u^2}} = \frac{1}{2} \int \frac{du}{\sqrt{1 - u^2}} $$ The integral $\int \frac{du}{\sqrt{1 - u^2}}$ is a standard integral, and its result is $\arcsin(u)$. So we have: $$ \frac{1}{2} \arcsin(u) = \frac{1}{2} \arcsin(2y) $$

4. Solve the right-hand side integral The integral on the right-hand side is straightforward: $$ \int dx = x + C $$ where $C$ is the constant of integration.

5. Combine the results Combining the results from both sides, we get: $$ \frac{1}{2} \arcsin(2y) = x + C $$

6. Apply the initial condition We are given that $y(0) = 0.75$. Plugging these values into the equation, we get: $$ \frac{1}{2} \arcsin(2(0.75)) = 0 + C $$ $$ \frac{1}{2} \arcsin(1.5) = C $$ However, $\arcsin(1.5)$ is not defined since the domain of $\arcsin(x)$ is $-1 \le x \le 1$. The initial condition is $y(0) = 0.75$, which means when $x=0$, $y=0.75$. $$ \frac{1}{2}\arcsin(2y) = x + C \ \frac{1}{2}\arcsin(2(0.75)) = 0 + C \ \frac{1}{2}\arcsin(1.5) = C $$ Since $\arcsin(1.5)$ is undefined, there is no solution to the initial value problem.