JEE Main 2002Annual, Paper 1MathematicsEasyDifferential Equations163× since 2002The solution of the equation d2ydx2=e−2x\,{{{d^2}y} \over {d{x^2}}} = {e^{ - 2x}}dx2d2y=e−2xAOption A. e−2x4{{{e^{ - 2x}}} \over 4}4e−2xBOption B. e−2x4+cx+d{{{e^{ - 2x}}} \over 4} + cx + d4e−2x+cx+dCOption C. 14e−2x+cx2+d{1 \over 4}{e^{ - 2x}} + c{x^2} + d41e−2x+cx2+dDOption D. 14e−4x+cx+d\,{1 \over 4}{e^{ - 4x}} + cx + d41e−4x+cx+dCheckSkip · show solution
