The respective expressions for complimentary function and particular integral part of the solution of the differential equation $$\frac{{{{\text{d}}^4}{\text{y}}}}{{{\text{d}}{{\text{x}}^4}}} + 3\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} = 108{{\text{x}}^2}$$ are
A. $$\left[ {{{\text{c}}_1} + {{\text{c}}_2}{\text{x}} + {{\text{c}}_3}\sin \sqrt {3{\text{x}}} + {{\text{c}}_4}\cos \sqrt {3{\text{x}}} } \right]{\text{ and }}\left[ {3{{\text{x}}^4} - 12{{\text{x}}^2} + {\text{c}}} \right]$$
B. $$\left[ {{{\text{c}}_2} + {{\text{c}}_3}\sin \sqrt {3{\text{x}}} + {{\text{c}}_4}\cos \sqrt {3{\text{x}}} } \right]{\text{ and }}\left[ {5{{\text{x}}^4} - 12{{\text{x}}^2} + {\text{c}}} \right]$$
C. $$\left[ {{{\text{c}}_1} + {{\text{c}}_3}\sin \sqrt {3{\text{x}}} + {{\text{c}}_4}\cos \sqrt {3{\text{x}}} } \right]{\text{ and }}\left[ {3{{\text{x}}^4} - 12{{\text{x}}^2} + {\text{c}}} \right]$$
D. $$\left[ {{{\text{c}}_1} + {{\text{c}}_2}{\text{x}} + {{\text{c}}_3}\sin \sqrt {3{\text{x}}} + {{\text{c}}_4}\cos \sqrt {3{\text{x}}} } \right]{\text{ and }}\left[ {5{{\text{x}}^4} - 12{{\text{x}}^2} + {\text{c}}} \right]$$
Answer: Option A
This Question Belongs to Engineering Maths >> Differential Equations
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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