Differential Equations
Linear Differential Equations
Grade 12
Question:
<p><strong>97.</strong> The general solution of the differential equation \((1 + \tan y)(dx - dy) + 2x\, dy = 0\) is:</p><p>[<strong>Note:</strong> Where \(C\) is constant of integration.]</p>
<p>\(x(\sin y + \cos y) = \sin y + Ce^{y}\)</p>
<p>\(x(\sin y + \cos y) = \sin y + Ce^{-y}\)</p>
<p>\(y(\sin x + \cos x) = \sin x + Ce^{x}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Rearrange the DE to recognize it as a linear first-order equation in x as a function of y, then apply the integrating factor method with the form dx/dy + P(y)·x = Q(y).
<p><strong>Step 1:</strong> Expand and rearrange the given equation (1 + tan y)(dx - dy) + 2x dy = 0:</p><p>(1 + tan y)dx − (1 + tan y)dy + 2x dy = 0</p><p>(1 + tan y)dx + [2x − (1 + tan y)]dy = 0</p><p><strong>Step 2:</strong> Rearrange as a linear DE in x with respect to y:</p><p>dx/dy + [2 − (1 + tan y)/(1 + tan y)]x = (1 + tan y)/(1 + tan y)</p><p>dx/dy + x/(1 + tan y) = 1</p><p><strong>Step 3:</strong> Rewrite: (1 + tan y)dx/dy + x = 1 + tan y</p><p>Or equivalently: dx/dy − x·(tan y)/(1 + tan y) = 1/(1 + tan y)</p><p><strong>Step 4:</strong> The integrating factor is μ(y) = e^{∫[−tan y/(1 + tan y)]dy} = e^{ln(cos y) − ln(1 + tan y)} = cos y/(1 + tan y)</p><p>Multiply through and recognize that d/dy[x·cos y/(1 + tan y)] = cos y/(1 + tan y)</p><p><strong>Step 5:</strong> Integrate: x·cos y/(1 + tan y) = ∫cos y/(1 + tan y) dy = sin y − cos y + C</p><p><strong>Step 6:</strong> Simplify using 1 + tan y = (cos y + sin y)/cos y:</p><p>x(cos y + sin y) = (sin y − cos y)cos y + C·cos y</p><p>∴ General solution: <strong>x = sin y cos y + C cos y/(sin y + cos y)</strong> or equivalent form <strong>x(sin y + cos y) − sin y cos y = C</strong></p>
Correct Answer: B