Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>If <em>y₁, y₂</em> are two solutions of the differential equation <em>dy/dx + P(x)·y = Q(x)</em>, then prove that <em>y = y₁ + c(y₁ - y₂)</em> is the general solution of the equation where <em>c</em> is any constant. For what relation between the constants <em>α, β</em> will the linear combination <em>αy₁ + βy₂</em> also be a solution?</p>
<p>\(\alpha + \beta = 0\)</p>
<p>\(\alpha + \beta = 1\)</p>
<p>\(\alpha - \beta = 1\)</p>
<p>\(\alpha \cdot \beta = 1\)</p>
Step-by-Step Solution
Key Concept: A linear differential equation is satisfied by y₁ and y₂ independently. Their linear combination αy₁ + βy₂ is a solution only if the non-homogeneous term Q(x) vanishes (α + β = 1 for the given form), because the non-homogeneous part doesn't scale linearly.
<p><strong>Step 1: Verify y₁ and y₂ are solutions</strong></p><p>Since y₁, y₂ satisfy the equation:<br/>dy₁/dx + P(x)y₁ = Q(x) ... (i)<br/>dy₂/dx + P(x)y₂ = Q(x) ... (ii)</p><p><strong>Step 2: Find difference y₁ - y₂</strong></p><p>Subtracting (ii) from (i):<br/>d(y₁ - y₂)/dx + P(x)(y₁ - y₂) = 0<br/>This is a homogeneous equation, so (y₁ - y₂) is a solution of the homogeneous form.</p><p><strong>Step 3: Prove y = y₁ + c(y₁ - y₂) is general solution</strong></p><p>dy/dx = dy₁/dx + c·d(y₁ - y₂)/dx<br/>dy/dx + P(x)y = dy₁/dx + P(x)y₁ + c[d(y₁ - y₂)/dx + P(x)(y₁ - y₂)]<br/>= Q(x) + c(0) = Q(x) ✓<br/>Since c is arbitrary constant, this is the general solution.</p><p><strong>Step 4: Condition for αy₁ + βy₂ to be a solution</strong></p><p>d(αy₁ + βy₂)/dx + P(x)(αy₁ + βy₂)<br/>= α[dy₁/dx + P(x)y₁] + β[dy₂/dx + P(x)y₂]<br/>= αQ(x) + βQ(x) = (α + β)Q(x)</p><p>For this to equal Q(x), we need: <strong>α + β = 1</strong></p><p>∴ Answer: The relation is <strong>α + β = 1</strong> (equivalently, β = 1 - α)</p>
Correct Answer: B