If roots of the auxiliary equation of $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + {\text{a}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + {\text{by}} = 0$$ are real and equal, the general solution of the differential equation is
A. $${\text{y}} = {{\text{c}}_1}{{\text{e}}^{ - \frac{{{\text{ax}}}}{2}}} + {{\text{c}}_2}{{\text{e}}^{\frac{{{\text{ax}}}}{2}}}$$
B. $${\text{y}} = \left( {{{\text{c}}_1} + {{\text{c}}_2}{\text{x}}} \right){{\text{e}}^{ - \frac{{{\text{ax}}}}{2}}}$$
C. $${\text{y}} = \left( {{{\text{c}}_1} + {{\text{c}}_2}\ln {\text{x}}} \right){{\text{e}}^{ - \frac{{{\text{ax}}}}{2}}}$$
D. $${\text{y}} = \left( {{{\text{c}}_1}\cos {\text{x}} + {{\text{c}}_2}\sin {\text{x}}} \right){{\text{e}}^{ - \frac{{{\text{ax}}}}{2}}}$$
Answer: Option B
A. $${\text{y}} = \left( {{{\text{C}}_1} - {{\text{C}}_2}{\text{x}}} \right){{\text{e}}^{\text{x}}} + {{\text{C}}_3}\cos {\text{x}} + {{\text{C}}_4}\sin {\text{x}}$$
B. $${\text{y}} = \left( {{{\text{C}}_1} + {{\text{C}}_2}{\text{x}}} \right){{\text{e}}^{\text{x}}} - {{\text{C}}_2}\cos {\text{x}} + {{\text{C}}_4}\sin {\text{x}}$$
C. $${\text{y}} = \left( {{{\text{C}}_1} + {{\text{C}}_2}{\text{x}}} \right){{\text{e}}^{\text{x}}} + {{\text{C}}_3}\cos {\text{x}} + {{\text{C}}_4}\sin {\text{x}}$$
D. $${\text{y}} = \left( {{{\text{C}}_1} + {{\text{C}}_2}{\text{x}}} \right){{\text{e}}^{\text{x}}} + {{\text{C}}_3}\cos {\text{x}} - {{\text{C}}_4}\sin {\text{x}}$$
A. $$\sqrt {1 - {{\text{x}}^2}} = {\text{c}}$$
B. $$\sqrt {1 - {{\text{y}}^2}} = {\text{c}}$$
C. $$\sqrt {1 - {{\text{x}}^2}} + \sqrt {1 - {{\text{y}}^2}} = {\text{c}}$$
D. $$\sqrt {1 + {{\text{x}}^2}} + \sqrt {1 + {{\text{y}}^2}} = {\text{c}}$$

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