A) \[x=k\left( 1+{{e}^{-y}} \right)\]
B) \[\text{x=k}\left( {{\text{e}}^{\text{y}}}\text{+}{{\text{e}}^{\text{-y}}}\text{-2} \right)\]
C) \[\text{x=k}\left( {{\text{e}}^{\text{-y}}}\text{-1} \right)\]
D) \[\text{x=k}\left( {{\text{e}}^{\text{y}}}\text{-1} \right)\]
Correct Answer: C
Solution :
Rewriting given differential equation as, \[t=n{{T}_{1/2}}=0.74\times 4.47\times {{10}^{8}}\] ....(1) where \[t=3.3\times {{10}^{8}}yr\] Its auxiliary equation is \[Y\left( n,\alpha \right)\]so that M = 1, -1 Hence, \[\alpha -\] where, \[_{Z}{{Y}^{A}}{{+}_{0}}{{n}^{1}}{{\to }_{3}}L{{i}^{7}}{{+}_{2}}H{{e}^{4}}\] are arbitrary constants. Now, \[\Rightarrow \] \[_{\text{5}}{{\text{Y}}^{\text{10}}}{{\text{=}}_{\text{5}}}{{\text{B}}^{\text{10}}}\] \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\] So, solution of Eq. (i) is \[\Rightarrow \] ...(ii) Given that x = 0, when y = 0 Hence, Eq. (ii) gives \[34=-\frac{13.6}{{{n}^{2}}}\] \[\Rightarrow \] \[{{n}^{2}}=4\] .....(iii) Multiplying both sides of Eq. (ii) by \[\Rightarrow \] we get \[x.{{e}^{-y}}={{C}_{1}}+{{C}_{2}}{{e}^{-2y}}-k{{e}^{-y}}\] Given that \[x\to m\]when \[y\to \infty ,\] m being a finite quantity. So, Eq. (iv) becomes \[x\times 0={{C}_{1}}+{{C}_{2}}\times 0-\left( k\times 0 \right)\] \[\Rightarrow \] \[{{C}_{1}}=0\] From Eqs. (iv) and (v), we get \[{{C}_{1}}=0\] and \[{{C}_{2}}=k\] Hence, Eq. (ii) becomes \[x=k{{e}^{-y}}-k\] \[=k\left( {{e}^{-y}}-1 \right)\]
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