Question

The equation of the curve passing through the point \( (0, \pi) \) and satisfying the differential equation \( ydx = (x + y^3 \cos y)dy \) is

• A. \( x = y^2 \sin y + y\cos^2 y \)
• B. \( x = y^2 \sin y + 2y\cos^2 \frac{y}{2} \)
• C. \( x = y^2 \sin y + y\cos^2 \frac{y}{2} \)
• D. \( x = y^2 \sin y - y\cos^2 y \)

Hint:

When solving differential equations, it's crucial to first identify their type (e.g., separable, exact, homogeneous, or linear). For first-order linear differential equations, which are of the form \( \frac{dx}{dy} + P(y)x = Q(y) \) or \( \frac{dy}{dx} + P(x)y = Q(x) \), the integrating factor method is essential. The integrating factor is calculated as \( IF = e^{\int P(y) dy} \) or \( IF = e^{\int P(x) dx} \), respectively. The general solution is then given by \( x \cdot IF = \int Q(y) \cdot IF dy + C \) or \( y \cdot IF = \int Q(x) \cdot IF dx + C \). After obtaining the general solution, use any given initial conditions (a point the curve passes through) to determine the value of the constant of integration. Finally, be prepared to use trigonometric identities to transform your solution into a form that matches the provided options.

Answer 1

The Correct Option is B

Step 1: Rearrange the differential equation into a standard form.
The given differential equation is \( ydx = (x + y^3 \cos y)dy \). We can rewrite this as: \[ y\frac{dx}{dy} = x + y^3 \cos y \] To put it in the standard form of a linear differential equation, \( \frac{dx}{dy} + P(y)x = Q(y) \): \[ y\frac{dx}{dy} - x = y^3 \cos y \] Divide by \(y\): \[ \frac{dx}{dy} - \frac{1}{y}x = y^2 \cos y \] This is a linear differential equation where \( P(y) = -\frac{1}{y} \) and \( Q(y) = y^2 \cos y \). 
Step 2: Find the integrating factor (IF).
The integrating factor is given by \( IF = e^{\int P(y) dy} \). \[ IF = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = e^{\ln|y|^{-1}} = e^{\ln\left(\frac{1}{|y|}\right)} = \frac{1}{y} \] (We assume \( y>0 \) since the point \( (0, \pi) \) implies \( y>0 \)). 
Step 3: Solve the differential equation using the integrating factor.
Multiply the standard form of the differential equation by the integrating factor: \[ \frac{1}{y}\left(\frac{dx}{dy} - \frac{1}{y}x\right) = \frac{1}{y}\left(y^2 \cos y\right) \] \[ \frac{1}{y}\frac{dx}{dy} - \frac{1}{y^2}x = y \cos y \] The left-hand side is the derivative of \( \left(x \cdot \frac{1}{y}\right) \) with respect to \(y\): \[ \frac{d}{dy}\left(\frac{x}{y}\right) = y \cos y \] Integrate both sides with respect to \(y\): \[ \int \frac{d}{dy}\left(\frac{x}{y}\right) dy = \int y \cos y dy \] \[ \frac{x}{y} = \int y \cos y dy \] To evaluate \( \int y \cos y dy \), we use integration by parts, \( \int u dv = uv - \int v du \).
Let \( u = y \) and \( dv = \cos y dy \).
Then \( du = dy \) and \( v = \sin y \).
\[ \int y \cos y dy = y \sin y - \int \sin y dy \] \[ = y \sin y - (-\cos y) + C_1 \] \[ = y \sin y + \cos y + C_1 \] So, the solution is: \[ \frac{x}{y} = y \sin y + \cos y + C_1 \] Multiply by \(y\) to find \(x\): \[ x = y^2 \sin y + y \cos y + C_1y \] 
Step 4: Use the given point to find the constant of integration C.
The curve passes through the point \( (0, \pi) \). Substitute \( x=0 \) and \( y=\pi \) into the equation: \[ 0 = (\pi)^2 \sin(\pi) + \pi \cos(\pi) + C_1\pi \] We know that \( \sin(\pi) = 0 \) and \( \cos(\pi) = -1 \). \[ 0 = \pi^2 (0) + \pi (-1) + C_1\pi \] \[ 0 = 0 - \pi + C_1\pi \] \[ 0 = \pi(C_1 - 1) \] Since \( \pi \neq 0 \), we must have \( C_1 - 1 = 0 \), which means \( C_1 = 1 \). 
Step 5: Write the final equation of the curve and match with options.
Substitute \( C_1 = 1 \) back into the equation for \(x\): \[ x = y^2 \sin y + y \cos y + y \] Now, let's compare this with the given options. Option (2) is \( x = y^2 \sin y + 2y\cos^2 \frac{y}{2} \).
We need to check if our solution can be transformed into option (2). Recall the half-angle identity for cosine: \( \cos y = 2\cos^2 \frac{y}{2} - 1 \).
Substitute this into our solution: \[ x = y^2 \sin y + y(2\cos^2 \frac{y}{2} - 1) + y \] \[ x = y^2 \sin y + 2y\cos^2 \frac{y}{2} - y + y \] \[ x = y^2 \sin y + 2y\cos^2 \frac{y}{2} \] This matches option (2).