The solution of the differential equation $$\frac{{{\text{dy}}}}{{{\text{dx}}}} + \frac{{\text{y}}}{{\text{x}}} = {\text{x}},$$ with the condition that y = 1 at x = 1, is
A. $${\text{y}} = \frac{2}{{3{{\text{x}}^2}}} + \frac{{\text{x}}}{3}$$
B. $${\text{y}} = \frac{{\text{x}}}{2} + \frac{{\text{1}}}{{2{\text{x}}}}$$
C. $${\text{y}} = \frac{2}{3} + \frac{{\text{x}}}{3}$$
D. $${\text{y}} = \frac{2}{{3{\text{x}}}} + \frac{{{{\text{x}}^2}}}{3}$$
Answer: Option D
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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