Consider the differential equation: $$\frac{{{\text{dy}}}}{{{\text{dx}}}} = \left( {1 + {{\text{y}}^2}} \right){\text{x}}{\text{.}}$$
The general solution with constant c is
A. $${\text{y}} = \tan \frac{{\text{x}}}{2} + \tan {\text{c}}$$
B. $${\text{y}} = {\tan ^2}\left( {\frac{{\text{x}}}{2} + {\text{c}}} \right)$$
C. $${\text{y}} = {\tan ^2}\left( {\frac{{\text{x}}}{2}} \right) + {\text{c}}$$
D. $${\text{y}} = \tan \left( {\frac{{{{\text{x}}^2}}}{2} + {\text{c}}} \right)$$
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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