Question

Y'(4+x²)=4+y², y(0)=π/2

Answer 1

To solve the given differential equation, we can follow these steps:

Step 1: Rewrite the equation in a more standard form. Step 2: Separate variables. Step 3: Integrate both sides. Step 4: Solve for the constant of integration using the initial condition. Step 5: Write the final solution.

Let's go through these steps one by one.

Step 1: Rewrite the equation in a more standard form. The given differential equation is: y'(4 + x²) = 4 + y²

Step 2: Separate variables. To separate variables, we'll move all the terms involving y to one side and all the terms involving x to the other side: y' = (4 + y²) / (4 + x²)

Step 3: Integrate both sides. Integrating both sides of the equation gives: ∫1/(4 + y²) dy = ∫1/(4 + x²) dx

Step 4: Solve for the constant of integration using the initial condition. To solve the integrals, we can use a trigonometric substitution. Let y = 2tan(theta), and dy = 2sec²(theta) d(theta). Substituting these values, we have: ∫1/(4 + 4tan²(theta)) * 2sec²(theta) d(theta) = ∫1/(4 + x²) dx

Simplifying the left side: ∫2sec²(theta)/(4 + 4tan²(theta)) d(theta) = ∫1/(4 + x²) dx

Simplifying the denominator on the left side: ∫sec²(theta)/(1 + tan²(theta)) d(theta) = ∫1/(4 + x²) dx

Using the identity: sec²(theta) = 1 + tan²(theta) ∫1 d(theta) = ∫1/(4 + x²) dx

Integrating both sides: θ = arctan(x/2) + C1

Step 5: Write the final solution. Now, we need to solve for y in terms of x using the initial condition y(0) = π/2. Substituting x = 0 into the equation above: θ = arctan(0/2) + C1 θ = 0 + C1 θ = C1

Using the initial condition y(0) = π/2: π/2 = C1

Substituting C1 back into the equation θ = arctan(x/2) + C1: θ = arctan(x/2) + π/2

Since y = 2tan(theta): y = 2tan(arctan(x/2) + π/2) y = 2tan(arctan(x/2)) + 2tan(π/2) y = 2(x/2) + ∞ (since tan(π/2) is undefined)

Therefore, the final solution to the given differential equation with the initial condition y(0) = π/2 is: y = x + ∞