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.
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