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 equation to separate variables by grouping dx and dy terms strategically, then recognize that d(xy) and d(sin y) patterns emerge when you divide by appropriate factors.
<p><strong>Step 1:</strong> Expand and rearrange the given equation:<br/>(1 + tan y)dx - (1 + tan y)dy + 2x dy = 0<br/>(1 + tan y)dx + [2x - (1 + tan y)]dy = 0<br/>(1 + tan y)dx + (2x - 1 - tan y)dy = 0</p><p><strong>Step 2:</strong> Rewrite strategically:<br/>(1 + tan y)dx + 2x dy - (1 + tan y)dy = 0<br/>Rearrange: (1 + tan y)dx + 2x dy = (1 + tan y)dy</p><p><strong>Step 3:</strong> Recognize that the left side is related to d(x(1 + tan y)). Divide the original equation by (1 + tan y):<br/>dx + (2x dy - dy)/(1 + tan y) = dy<br/>Or work with: d[x(1 + tan y)] = d(x + x tan y)</p><p><strong>Step 4:</strong> Rewrite as:<br/>dx + 2x dy - dy - tan y dy = 0<br/>dx - tan y dy + d(2xy) - 2x dy = 0<br/>Rearrange: dx + 2x dy = dy + tan y dy<br/>d(x) + d(2xy) - 2x dy = sin y/cos y · dy<br/>Simplify: d[x + 2xy] = d(-ln|cos y|)</p><p><strong>Step 5:</strong> Integrate both sides:<br/>x(1 + 2y) = -ln|cos y| + C<br/>Or equivalently: <strong>x(1 + 2y) + ln|cos y| = C</strong></p><p>∴ Answer: B</p>
Correct Answer: B