The solution of $${\text{x}}\frac{{{\text{dy}}}}{{{\text{dx}}}} + {\text{y}} = {{\text{x}}^4}$$ with the condition $${\text{y}}\left( 1 \right) = \frac{6}{5}$$ is
A. $${\text{y}} = \frac{{{{\text{x}}^4}}}{5} + \frac{1}{{\text{x}}}$$
B. $${\text{y}} = \frac{{4{{\text{x}}^4}}}{5} + \frac{4}{{5{\text{x}}}}$$
C. $${\text{y}} = \frac{{{{\text{x}}^4}}}{5} + 1$$
D. $${\text{y}} = \frac{{{{\text{x}}^5}}}{5} + 1$$
Answer: Option A
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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