Differential Equations
Particular solution of linear ODE
Grade Class 12
Question:
<p>The value of the constants \(m\) and \(c\) for which \(y = mx + c\) is a solution of the differential equation \(2y'' - 5y' - 4y = -4x\) is:</p>
<span>\(m=1,\ c=5/4\)</span>
<span>\(m=1,\ c=-5/4\)</span>
<span>\(\text{No such real }m,c\text{ exist}\)</span>
<span>\(m=-1,\ c=3/4\)</span>
Step-by-Step Solution
Key Concept: Substitute y = mx + c into the ODE and compare coefficients.
<div class='solution'><p><strong>Step 1:</strong> For \(y = mx+c\): \(y' = m\), \(y'' = 0\).</p>
<p><strong>Step 2:</strong> Substituting into \(2y'' - 5y' - 4y = -4x\):</p>
<p>\[2(0) - 5m - 4(mx+c) = -4x\]</p>
<p>\[-4mx - (5m+4c) = -4x\]</p>
<p><strong>Step 3:</strong> Comparing coefficients of \(x\): \(-4m = -4 \Rightarrow m = 1\).</p>
<p><strong>Step 4:</strong> Comparing constants: \(-(5m+4c) = 0 \Rightarrow 5 + 4c = 0 \Rightarrow c = -\dfrac{5}{4}\).</p>
<p><strong>Answer: (B)</strong> \(m=1,\ c = -\dfrac{5}{4}\)</p>
<p class='key-concept'>🔑 Key Concept: For a particular solution of the form \(y = mx+c\), all derivatives of order ≥ 2 vanish. Coefficient matching is the only tool needed.</p>
<p class='trap-warning'>⚠️ Trap Warning: Students often forget to zero out \(y''\) and try to match without substituting properly, getting \(c = 5/4\) instead of \(-5/4\).</p></div>
Correct Answer: 2