Consider a system governed by the following equations:
$$\frac{{{\text{d}}{{\text{x}}_1}\left( {\text{t}} \right)}}{{{\text{dt}}}} = {{\text{x}}_2}\left( {\text{t}} \right) - {{\text{x}}_1}\left( {\text{t}} \right);\,\frac{{{\text{d}}{{\text{x}}_2}\left( {\text{t}} \right)}}{{{\text{dt}}}} = {{\text{x}}_1}\left( {\text{t}} \right) - {{\text{x}}_2}\left( {\text{t}} \right)$$
The initial conditions are such that $${{\text{x}}_1}\left( 0 \right) < {{\text{x}}_2}\left( 0 \right) < \infty .$$ Let $${{\text{x}}_{1{\text{f}}}} = \mathop {\lim }\limits_{{\text{t}} \to \infty } {{\text{x}}_1}\left( {\text{t}} \right)$$ and $${{\text{x}}_{2{\text{f}}}} = \mathop {\lim }\limits_{{\text{t}} \to \infty } {{\text{x}}_2}\left( {\text{t}} \right).$$ Which one of the following is true?
A. $${{\text{x}}_{1{\text{f}}}} < {{\text{x}}_{2{\text{f}}}} < \infty $$
B. $${{\text{x}}_{2{\text{f}}}} < {{\text{x}}_{1{\text{f}}}} < \infty $$
C. $${{\text{x}}_{1{\text{f}}}} = {{\text{x}}_{2{\text{f}}}} < \infty $$
D. $${{\text{x}}_{1{\text{f}}}} = {{\text{x}}_{2{\text{f}}}} = \infty $$
Answer: Option C
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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