The differential equation corresponding to the curve \( y=e^{-2x}(ax-b) \), where \( a \) & \( b \) are arbitrary constants, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( y''+4y'+4y=0 \)
There are two arbitrary constants, so the equation comes from differentiating twice and eliminating them. Moving \( e^{-2x} \) to the left:
\( \displaystyle ye^{2x}=ax-b \)
Differentiating (product rule on the left):
\( \displaystyle e^{2x}(y'+2y)=a \)
Differentiating again, since a is a constant:
\( \displaystyle e^{2x}(y''+2y')+2e^{2x}(y'+2y)=0 \)
Dividing by \( e^{2x}\ne 0 \):
\( \displaystyle y''+2y'+2y'+4y=0 \)
\( \displaystyle y''+4y'+4y=0 \)