Question:
The function \(y = be^x + ae^{-x}\), \(a\) and \(b\) are constants, is a solution of
The function \(y = be^x + ae^{-x}\), \(a\) and \(b\) are constants, is a solution of
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Functions of the form $ae^{x} + be^{-x}$:
- Usually satisfy equations like $y'' - y = 0$
- Always check by direct substitution
Updated On: Apr 30, 2026
- $y'' + y = 0$
- $y'' - y = 0$
- $y'' + x = 0$
- $y'' - 2y = 0$
- $y'' + xy = 0$
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Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
To verify a function is a solution of a differential equation:
• Find derivatives
• Substitute into the equation
• Check if LHS = RHS
Step 1: Given function.
\[ y = be^x + ae^{-x} \]
Step 2: Find first derivative.
\[ y' = be^x - ae^{-x} \]
Step 3: Find second derivative.
\[ y'' = be^x + ae^{-x} \]
Step 4: Substitute into $y'' - y$.
\[ y'' - y = (be^x + ae^{-x}) - (be^x + ae^{-x}) \] \[ = 0 \]
Step 5: Verify equation.
\[ y'' - y = 0 \]
• Find derivatives
• Substitute into the equation
• Check if LHS = RHS
Step 1: Given function.
\[ y = be^x + ae^{-x} \]
Step 2: Find first derivative.
\[ y' = be^x - ae^{-x} \]
Step 3: Find second derivative.
\[ y'' = be^x + ae^{-x} \]
Step 4: Substitute into $y'' - y$.
\[ y'' - y = (be^x + ae^{-x}) - (be^x + ae^{-x}) \] \[ = 0 \]
Step 5: Verify equation.
\[ y'' - y = 0 \]
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