Differential Equations
Formation of differential equations
Grade 12
Question:
<p>We have \(y = ae^{3x} + be^x\). Eliminating \(a\) and \(b\), we get a differential equation of the form \(\dfrac{d^2y}{dx^2} - 4\dfrac{dy}{dx} + 3y = 0\) with \(a = 1,\, b = -4,\, c = 3\). Find \(a + b + c\).</p>
<p>0</p>
<p>1</p>
<p>-1</p>
<p>2</p>
Step-by-Step Solution
Key Concept: To eliminate arbitrary constants from a solution involving two parameters, differentiate twice and form a system of equations. The coefficients in the resulting differential equation come directly from matching terms after substitution.
<p><strong>Step 1:</strong> Start with the general solution: y = ae^(3x) + be^x</p><p><strong>Step 2:</strong> Differentiate once: dy/dx = 3ae^(3x) + be^x</p><p><strong>Step 3:</strong> Differentiate again: d²y/dx² = 9ae^(3x) + be^x</p><p><strong>Step 4:</strong> From the expressions above, notice that:</p><ul><li>y = ae^(3x) + be^x</li><li>dy/dx = 3ae^(3x) + be^x</li><li>d²y/dx² = 9ae^(3x) + be^x</li></ul><p><strong>Step 5:</strong> Eliminate ae^(3x) and be^x:</p><p>From dy/dx: dy/dx - y = 2ae^(3x)</p><p>From d²y/dx²: d²y/dx² - dy/dx = 6ae^(3x)</p><p>From d²y/dx² - 4(dy/dx) + 3y: Combining strategically gives d²y/dx² - 4(dy/dx) + 3y = 0</p><p><strong>Step 6:</strong> The differential equation is d²y/dx² - 4(dy/dx) + 3y = 0</p><p>Comparing with the form where coefficients are a = 1, b = -4, c = 3</p><p>∴ a + b + c = 1 + (-4) + 3 = <strong>0</strong></p>
Correct Answer: A