Differential Equations
Differential Equation of Family of Curves
Grade 12
Question:
<p>Let \(y = (a\sin x + (b+c)\cos x)e^x + d\), where a, b, c and d are parameters represent a family of curves. The differential equation for the given family of curves is \(y'' - ay' + by = 0\). Then find \(a + b\).</p>
Step-by-Step Solution
Key Concept: Eliminate parameters by successive differentiation and match coefficients with the given differential equation form.
<p><strong>Step 1:</strong> Given $y = (a\sin x + (b+c)\cos x)e^x + d$</p><p><strong>Step 2:</strong> Find the first derivative: $y' = (a\sin x + (b+c)\cos x)e^x + (a\cos x - (b+c)\sin x)e^x$</p><p><strong>Step 3:</strong> Simplify: $y' = ((a + b + c)\sin x + (b + c - a)\cos x)e^x$</p><p><strong>Step 4:</strong> Find the second derivative similarly</p><p><strong>Step 5:</strong> The homogeneous part (without d) satisfies: $y'' - 2y' + 2y = 0$ (by characteristic equation analysis)</p><p><strong>Step 6:</strong> Comparing with $y'' - ay' + by = 0$, we get $a = 2$ and $b = 2$</p><p><strong>Step 7:</strong> Therefore $a + b = 4$</p>
Correct Answer: 4