Differential Equations
Solution of Differential Equations
Grade 12
Question:
<p>Given that the differential equation is \(\dfrac{d^2y}{dx^2} = e^{-2x}\). The general solution is:</p>
<p>(1) \(y = \dfrac{1}{4}e^{-2x} + cx + d\)</p>
<p>(2) \(y = \dfrac{1}{4}e^{-2x} + cx^2 + d\)</p>
<p>(3) \(y = -\dfrac{1}{4}e^{-2x} + cx + d\)</p>
<p>(4) \(y = \dfrac{1}{4}e^{2x} + cx + d\)</p>
Step-by-Step Solution
Key Concept: Integrate twice successively to find the general solution of a second-order ODE. Each integration introduces an arbitrary constant, and the first integral gives an intermediate function before finding y itself.
<p><strong>Step 1:</strong> Integrate the given equation with respect to x once:</p><p>∫(d²y/dx²)dx = ∫e^(-2x)dx</p><p>dy/dx = -½e^(-2x) + C₁</p><p><strong>Step 2:</strong> Integrate again with respect to x:</p><p>∫(dy/dx)dx = ∫[-½e^(-2x) + C₁]dx</p><p>y = -½ · (-½)e^(-2x) + C₁x + C₂</p><p>y = ¼e^(-2x) + C₁x + C₂</p><p><strong>Step 3:</strong> Verify by taking second derivative:</p><p>dy/dx = -½e^(-2x) + C₁</p><p>d²y/dx² = e^(-2x) ✓</p><p><strong>∴ General Solution:</strong> y = ¼e^(-2x) + C₁x + C₂ (where C₁, C₂ are arbitrary constants)</p>
Correct Answer: A