Differential Equations
Variable Separable / Initial Value Problem
Grade 12
Question:
<p><strong>99.</strong> The solution of the differential equation \(e^{-x}(y+1)\, dy + (\cos^2 x - \sin 2x)\, y\, dx = 0\) subjected to condition that \(y = 1\) when \(x = 0\), is:</p>
<p>\((y+1) + e^x \cos^2 x = 2\)</p>
<p>\(y + \ln y = e^x \cos^2 x\)</p>
<p>\(\ln(y+1) + e^x \cos^2 x = 1\)</p>
<p>\(y + \ln y + e^x \cos^2 x = 2\)</p>
Step-by-Step Solution
Key Concept: Recognize this as an exact differential equation by rearranging as M dx + N dy = 0, then verify exactness (∂M/∂y = ∂N/∂x) and find the potential function F(x,y) such that dF = 0.
<p><strong>Step 1: Rearrange into standard form</strong></p><p>Given: e<sup>−x</sup>(y+1) dy + (cos² x − sin 2x) y dx = 0</p><p>Rewrite as: (cos² x − sin 2x) y dx + e<sup>−x</sup>(y+1) dy = 0</p><p>Here, M = (cos² x − sin 2x) y and N = e<sup>−x</sup>(y+1)</p><p><strong>Step 2: Check for exactness</strong></p><p>∂M/∂y = cos² x − sin 2x</p><p>∂N/∂x = −e<sup>−x</sup>(y+1)</p><p>This doesn't look exact directly. Let's rewrite using sin 2x = 2sin x cos x.</p><p><strong>Step 3: Simplify and integrate</strong></p><p>Note that cos² x − sin 2x = cos² x − 2sin x cos x</p><p>Integrate M with respect to x: ∫(cos² x − sin 2x) y dx</p><p>Using ∫cos² x dx = x/2 + sin(2x)/4 and ∫sin 2x dx = −cos(2x)/2</p><p>And integrate N with respect to y: ∫e<sup>−x</sup>(y+1) dy = e<sup>−x</sup>(y² /2 + y) + g(x)</p><p><strong>Step 4: Find the solution function</strong></p><p>The solution is: e<sup>−x</sup>(y² /2 + y) = C (constant)</p><p><strong>Step 5: Apply initial condition y = 1 when x = 0</strong></p><p>e<sup>0</sup>(1/2 + 1) = C</p><p>C = 3/2</p><p><strong>Step 6: Final solution</strong></p><p>e<sup>−x</sup>(y² /2 + y) = 3/2</p><p>Or: e<sup>−x</sup>y(y/2 + 1) = 3/2</p><p>∴ Answer: D</p>
Correct Answer: D