ExamVeda
Login
Home

The general solution of the differential equation frac text d ^4...

This question belongs to Engineering Maths Differential Equations
Differential Equations
?

The general solution of the differential equation, $$\frac{{{{\text{d}}^4}{\text{y}}}}{{{\text{d}}{{\text{x}}^4}}} - 2\frac{{{{\text{d}}^3}{\text{y}}}}{{{\text{d}}{{\text{x}}^3}}} + 2\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - 2\frac{{{\text{dy}}}}{{{\text{dx}}}} + {\text{y}} = 0$$       is

Answer & Solution
Correct Answer: Option C
Let's discuss the solution. Join the discussion
Examveda
Question posted by Examveda
Community

Join the Discussion

No comments yet Be the first to discuss this question.
Keep Learning

Related Questions on Differential Equations

The general solution of the differential equation, $$\frac{{{{\text{d}}^4}{\text{y}}}}{{{\text{d}}{{\text{x}}^4}}} - 2\frac{{{{\text{d}}^3}{\text{y}}}}{{{\text{d}}{{\text{x}}^3}}} + 2\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{x}}^2}}} - 2\frac{{{\text{dy}}}}{{{\text{dx}}}} + {\text{y}} = 0$$       is
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. $${\text{y}} = \left( {{{\text{C}}_1} + {{\text{C}}_2}{\text{x}}} \right){{\text{e}}^{\text{x}}} + {{\text{C}}_3}\cos {\text{x}} - {{\text{C}}_4}\sin {\text{x}}$$
The solution of the differential equation, $${\text{y}}\sqrt {1 - {{\text{x}}^2}} {\text{dy}} + {\text{x}}\sqrt {1 - {{\text{y}}^2}} {\text{dx}} = {\text{0}}$$      is
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. $$\sqrt {1 + {{\text{x}}^2}} + \sqrt {1 + {{\text{y}}^2}} = {\text{c}}$$
The maximum value of the solution y(t) of the differential equation \[{\rm{y}}\left( {\rm{t}} \right) + {\rm{\ddot y}}\left( {\rm{t}} \right) = 0\]    with initial conditions \[{\rm{\dot y}}\left( 0 \right) = 1\]  and y(0) = 1, for t ≥ 0 is
A. 1 B. 2 C. $$\pi $$ D. $$\sqrt 2 $$
The order of the differential equation $$\frac{{{{\text{d}}^2}{\text{y}}}}{{{\text{d}}{{\text{t}}^2}}} + {\left( {\frac{{{\text{dy}}}}{{{\text{dt}}}}} \right)^3} + {{\text{y}}^4} = {{\text{e}}^{ - {\text{t}}}}$$     is
A. 1 B. 2 C. 3 D. 4
More Related Questions on Differential Equations