For $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} + 4\frac{{{\text{dy}}}}{{{\text{dx}}}} + 3{\text{y}} = 3{{\text{e}}^{2{\text{x}}}},$$ the particular integral is
A. $$\frac{1}{{15}}{{\text{e}}^{2{\text{x}}}}$$
B. $$\frac{1}{5}{{\text{e}}^{2{\text{x}}}}$$
C. 3e2x
D. C1e-x + C2e-3x
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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