Verified Solution • 2025–26 Session moderate

QJEEM18Madef2: JEE Main 2018 Mathematics - Class 12 JEEM18Madef2 Solution: Let y = y(x) be the solution of the differential equation where …

Let y = y(x) be the solution of the differential equation dydx+2y=f(x),{{dy} \over {dx}} + 2y = f\left( x \right),

where f(x)={1,x∈[0,1]0,otherwise

If y(0) = 0, then y(32)y\left( {{3 \over 2}} \right) is :

Multiple Choice Options

  • A e2+12e4{{{e^2} + 1} \over {2{e^4}}}
  • B 12e{1 \over {2e}}
  • C e2−1e3{{{e^2} - 1} \over {{e^3}}}
  • D e2−12e3{{{e^2} - 1} \over {2{e^3}}}

Step-by-Step Verified Solution

When x ∈\in [0, 1], then dydx{{dy} \over {dx}} + 2y = 1 

⇒\Rightarrow  y = 12{1 \over 2} + C1e −- 2x

∵\because y(0) = 0 ⇒\Rightarrow y(x) = 12{1 \over 2} −- 12{1 \over 2} e −- 2x

Here, y(1) = 12{1 \over 2} −- 12{1 \over 2} e −- 2 = e2−12e2{{{e^2} - 1} \over {2{e^2}}}

When x∉[0,1]x \notin \left[ {0,1} \right] , then dydx{{dy} \over {dx}} + 2y = 0 ⇒\Rightarrow y = c2 e −- 2x

∵\because   y(1) = e2−12{{{e^2} - 1} \over 2} ⇒\Rightarrow e2−12{{{e^2} - 1} \over 2} = c2e −- 2 ⇒\Rightarrow  C2 = e2−12{{{e^2} - 1} \over 2}

∴\therefore   y(x) (e2−12)e−2x⇒y(32)\left( {{{{e^2} - 1} \over 2}} \right){e^{ - 2x}} \Rightarrow y\left( {{3 \over 2}} \right) = e2−12e3{{{e^2} - 1} \over {2{e^3}}}

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