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Differential Equations question

2002 · Shift 0 · Q85

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Differential Equations
2002 · Shift 0 · Q85

Differential Equations question

2002 · Shift 0 · Q85

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

The solution of the equation

\,{{{d^2}y} \over {d{x^2}}} = {e^{ - 2x}}

A
${{{e^{ - 2x}}} \over 4}$
B
${{{e^{ - 2x}}} \over 4} + cx + d$
C
${1 \over 4}{e^{ - 2x}} + c{x^2} + d$
D
$\,{1 \over 4}{e^{ - 4x}} + cx + d$

View written solutionFree

Correct answer: B

We need to solve the differential equation $\frac{d^2y}{dx^2}=e^{-2x}.$
Integrate once with respect to $x$ : $\frac{dy}{dx}=\int e^{-2x}\,dx = -\frac{1}{2}e^{-2x}+C_1.$
Integrate again: $y=\int\left(-\frac{1}{2}e^{-2x}+C_1\right)dx.$

Now, $\int -\frac{1}{2}e^{-2x}dx=\frac{1}{4}e^{-2x}$ because $\frac{d}{dx}\left(\frac{1}{4}e^{-2x}\right)=-\frac{1}{2}e^{-2x}.$

Also, $\int C_1\,dx=C_1x.$

So, $y=\frac{1}{4}e^{-2x}+C_1x+C_2.$

Compare with the given options:

A: $\frac{e^{-2x}}{4}$ — missing the two arbitrary constants, so incorrect.
B: $\frac{e^{-2x}}{4}+cx+d$ — matches the general solution, so correct.
C: $\frac{1}{4}e^{-2x}+cx^2+d$ — $cx^2$ is incorrect.
D: $\frac{1}{4}e^{-4x}+cx+d$ — exponential term is incorrect.

Therefore, the correct option is $\boxed{B}.$

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