Differential Equations
Variable Separable / First Order
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 a separable differential equation by rearranging terms into the form dy/dx = f(x)g(y), then integrate both sides with respect to their respective variables using the initial condition to find the constant.
<p><strong>Step 1:</strong> Rearrange the given equation:</p><p>e^(-x)(y+1)dy + (cos²x - sin2x)y dx = 0</p><p>e^(-x)(y+1)dy = -(cos²x - sin2x)y dx</p><p><strong>Step 2:</strong> Separate variables:</p><p>(y+1)/y dy = -(cos²x - sin2x)e^x dx</p><p>(1 + 1/y)dy = -(cos²x - sin2x)e^x dx</p><p><strong>Step 3:</strong> Integrate both sides:</p><p>∫(1 + 1/y)dy = -∫(cos²x - sin2x)e^x dx</p><p>y + ln|y| = -∫cos²x·e^x dx + ∫sin2x·e^x dx + C</p><p><strong>Step 4:</strong> Use integration by parts and trigonometric identities on the right side. Note that cos²x = (1+cos2x)/2 and sin2x = 2sinxcosx.</p><p>After integration: y + ln|y| = -e^x·sinx·cosx + e^x·sinx + C</p><p>Or equivalently: y + ln|y| = e^x·sinx(1 - cosx) + C</p><p><strong>Step 5:</strong> Apply initial condition y=1 when x=0:</p><p>1 + ln(1) = e^0·sin(0)(1 - cos(0)) + C</p><p>1 + 0 = 0 + C</p><p>C = 1</p><p><strong>Step 6:</strong> Therefore, the solution is:</p><p>y + ln|y| = e^x·sinx(1 - cosx) + 1</p><p>∴ Answer: D</p>
Correct Answer: D