Question

If the differential equation of the family of curves given by \[ y=ae^x+b\cos x, \] where \(a\) and \(b\) are arbitrary constants, is \[ y_2(\cos x+\sin x)+y(\cos x-\sin x)=2y_1f(x), \] then \(f(x)=\)

• A. \(\sin x\)
• B. \(\cos x\)
• C. \(-\cos x\)
• D. \(-\sin x\)

Hint:

When a family of curves contains arbitrary constants, differentiate enough times to obtain relations involving \(y\), \(y'\), and \(y''\). Then eliminate the constants by substitution and simplification.

Answer 1

The Correct Option is B

Concept: To obtain the differential equation of a family containing two arbitrary constants, differentiate twice and eliminate the constants. Here, \[ y_1=\frac{dy}{dx}, \qquad y_2=\frac{d^2y}{dx^2}. \]

Step 1: Differentiate the given family. Given, \[ y=ae^x+b\cos x. \] Differentiating, \[ y_1=ae^x-b\sin x. \] Differentiating again, \[ y_2=ae^x-b\cos x. \]

Step 2: Evaluate the left-hand side. \[ y_2(\cos x+\sin x)+y(\cos x-\sin x). \] Substituting \(y\) and \(y_2\), \[ =(ae^x-b\cos x)(\cos x+\sin x) +(ae^x+b\cos x)(\cos x-\sin x). \] Collecting terms, \[ =ae^x\big[(\cos x+\sin x)+(\cos x-\sin x)\big] \] \[ \quad +b\cos x\big[-(\cos x+\sin x)+(\cos x-\sin x)\big]. \] \[ =2ae^x\cos x-2b\sin x\cos x. \] \[ =2\cos x\,(ae^x-b\sin x). \] Using \[ y_1=ae^x-b\sin x, \] \[ y_2(\cos x+\sin x)+y(\cos x-\sin x) = 2y_1\cos x. \]

Step 3: Compare with the given differential equation. Given, \[ y_2(\cos x+\sin x)+y(\cos x-\sin x) = 2y_1f(x). \] Comparing, \[ f(x)=\cos x. \] Therefore, \[ \boxed{f(x)=\cos x} \] \[ \boxed{\text{Answer = (B)}} \]