Question:
The differential equation whose general solution is \( y = e^x(A\cos x + B\sin x) \) is
The differential equation whose general solution is \( y = e^x(A\cos x + B\sin x) \) is
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General solution with \(e^{ax}(\cos bx,\sin bx)\) always corresponds to roots \(a \pm ib\).
Updated On: May 1, 2026
- \( y'' - 2y' + 2y = 0 \)
- \( y'' - 2y' - 2y = 0 \)
- \( y'' + 2y' + 2y = 0 \)
- \( y'' + 2y' - 2y = 0 \)
- \( y'' - 2y' + y = 0 \)
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The Correct Option is A
Solution and Explanation
Concept: If the general solution is given, we can reverse-engineer the differential equation using the characteristic (auxiliary) equation.
Step 1: Observe the structure: \[ y = e^x(A\cos x + B\sin x) \] This is a standard form: \[ e^{\alpha x}(C\cos \beta x + D\sin \beta x) \] which corresponds to roots: \[ \alpha \pm i\beta \]
Step 2: Identify parameters: \[ \alpha = 1,\quad \beta = 1 \] So roots are: \[ m = 1 \pm i \]
Step 3: Form auxiliary equation: \[ (m-(1+i))(m-(1-i))=0 \]
Step 4: Expand carefully: \[ = (m-1-i)(m-1+i) = (m-1)^2 + 1 \]
Step 5: Expand fully: \[ (m-1)^2 + 1 = m^2 - 2m + 1 + 1 = m^2 - 2m + 2 \]
Step 6: Replace \(m\) with differential operator: \[ D^2 - 2D + 2 = 0 \]
Step 7: Convert into differential equation: \[ y'' - 2y' + 2y = 0 \]
Step 1: Observe the structure: \[ y = e^x(A\cos x + B\sin x) \] This is a standard form: \[ e^{\alpha x}(C\cos \beta x + D\sin \beta x) \] which corresponds to roots: \[ \alpha \pm i\beta \]
Step 2: Identify parameters: \[ \alpha = 1,\quad \beta = 1 \] So roots are: \[ m = 1 \pm i \]
Step 3: Form auxiliary equation: \[ (m-(1+i))(m-(1-i))=0 \]
Step 4: Expand carefully: \[ = (m-1-i)(m-1+i) = (m-1)^2 + 1 \]
Step 5: Expand fully: \[ (m-1)^2 + 1 = m^2 - 2m + 1 + 1 = m^2 - 2m + 2 \]
Step 6: Replace \(m\) with differential operator: \[ D^2 - 2D + 2 = 0 \]
Step 7: Convert into differential equation: \[ y'' - 2y' + 2y = 0 \]
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