Question

The solution of the differential equation $(2 \cos y) y' + \sin y = x^2 \csc y, y \neq 0$ is:

• A. $\sin^2 y - (x - 1)^2 = 1 + c e^{-x}$; $c$ is arbitrary constant
• B. $\sin^2 y + (x - 1)^2 = c e^{x}$; $c$ is arbitrary constant
• C. $\sin^2 y = (x^2 + 2x + 2) + c e^{-x}$; $c$ is arbitrary constant
• D. $\sin^2 y = (x^2 - 2x + 2) + c e^{+x}$; $c$ is arbitrary constant

Hint:

Notice that $2 \sin y \cos y = \frac{d}{dx}(\sin^2 y)$. Seeing derivatives of composite functions like $\frac{d}{dx}(\sin^2 y)$ instantly reveals the correct variable substitution!

Answer 1

The Correct Option is A

Step 1: Concept
This non-linear differential equation can be transformed into a linear first-order differential equation using a suitable substitution (Bernoulli-type transformation).

Step 2: Key Formulas and Approach
1. Rearrange $2 \cos y \frac{dy}{dx} + \sin y = \frac{x^2}{\sin y}$. 2. Multiply by $\sin y$: \[ (2 \sin y \cos y) \frac{dy}{dx} + \sin^2 y = x^2 \] 3. Substitute $v = \sin^2 y \implies \frac{dv}{dx} = 2 \sin y \cos y \frac{dy}{dx}$. 4. Solve the resulting first-order linear ODE in $v$: $\frac{dv}{dx} + P(x)v = Q(x)$ using Integrating Factor $I.F. = e^{\int P(x) dx}$.

Step 3: Step-by-step Explanation

• Substitute $v = \sin^2 y$: \[ \frac{dv}{dx} + v = x^2 \]
• Compute Integrating Factor: \[ I.F. = e^{\int 1 dx} = e^x \]
• Multiply by $I.F.$ and integrate: \[ v \cdot e^x = \int x^2 e^x dx + c \]
• Integrate $\int x^2 e^x dx$ using integration by parts: \[ \int x^2 e^x dx = x^2 e^x - 2x e^x + 2e^x = e^x (x^2 - 2x + 2) \]
• Substitute back into solution equation: \[ v \cdot e^x = e^x (x^2 - 2x + 2) + c \]
• Divide through by $e^x$: \[ v = x^2 - 2x + 2 + c e^{-x} \]
• Substitute $v = \sin^2 y$: \[ \sin^2 y = x^2 - 2x + 2 + c e^{-x} \]
• Rewrite $x^2 - 2x + 2$ as $(x - 1)^2 + 1$: \[ \sin^2 y = (x - 1)^2 + 1 + c e^{-x} \] \[ \sin^2 y - (x - 1)^2 = 1 + c e^{-x} \]

Step 4: Final Answer
The general solution is $\sin^2 y - (x - 1)^2 = 1 + c e^{-x}$. Thus, Option (A) is correct.