JEE Main & Advanced JEE Main Online Paper (Held On 15 April 2018) Slot-I

  • question_answer Let \[y=y(x)\] be the solution of the differential equation \[\frac{dy}{dx}+2y=f(x)\], where \[f(x)=\left\{ \begin{matrix}    1, & x\in [0,1]  \\    0, & otherwise  \\ \end{matrix} \right.\] If \[\text{y(0)=0,then}\,\,\,\text{y}\,\,\left( \frac{\text{3}}{\text{2}} \right)\text{is}\]                          [JEE Online 15-04-2018]

    A) \[\frac{{{e}^{2}}-1}{2{{e}^{3}}}\]       

    B) \[\frac{{{e}^{2}}-1}{{{e}^{3}}}\]         

    C) \[\frac{1}{2e}\]                                  

    D) \[\frac{{{e}^{2}}+1}{2{{e}^{4}}}\]     

    Correct Answer: A

    Solution :

    Solving the initial value problem, we get \[y=\frac{1}{2}-\frac{1}{2}{{e}^{-2x}}\]when \[x\in [0,1]\]. We can check this by substituting this in the differential equation and checking the initial value. So,\[y(1)=\frac{1-{{e}^{-2}}}{2}=\frac{{{e}^{2}}-1}{2{{e}^{2}}}.......(1)\] Now, for\[x\in (1,\infty ),\] we have \[{{e}^{2x}}y={{c}_{2}}\](solving the differential equation separately for this interval) Using the condition found above in (1) , we have  \[{{c}_{2}}=\frac{{{e}^{2}}-1}{2}\]. That gives \[y=\frac{{{e}^{2}}-1}{2}{{e}^{-2x}}\]for \[x\in (1,\infty )\] So, for \[x=\frac{3}{2}\], we get \[y=\frac{{{e}^{2}}-1}{2{{e}^{3}}}\]. So, the correct answer is option A.


You need to login to perform this action.
You will be redirected in 3 sec spinner