Differential Equation - Engineering Mathematics MCQ

Sol.[A]   The parametric form of the given equation is x = t, y = t². The equation of any tangent at
t is 2xt = y t². Differentiating, we get 2t = y₁. Putting this value in the equation of tangent, we have 2 x y₁/2 = y (y₁/2)² Þ 4xy₁ = 4y y₁²

                The order of this equation is one.

x = tan t + cot t

∴ x = [(sin²t + cos²t)/(sint cost)] = [2/(sin 2t)]

also

 y = 2 log (cot t)

(dx/dt) = (d/dt) [2/(sin 2t)]

(dx/dt) = – 2 ∙ cosec 2t ∙ cot 2t (2)

(dy/dt) = [2/{(cos t)/(sin t)}] ∙ (– cosec² t)

= [(– 2)/(cos t)] ∙ [(sin t)/(sin² t)] = [(– 4)/(sin 2t)]

∴ [(dy/dt)/(dx/dt)] = (dy/dx) = [(– 4 cosec 2t)/(– 4 cosec 2t ∙ cot 2t)]

∴ (dy/dx) = tan 2t

y = acosx + bsinx + 4

⟹dy/dx ​= −asinx + bcosx 

⟹d²y​/dx² = −acosx − bsinx 

⟹d²y​/dx² = −y + 4 

⟹ d²y​/dx² + y = 4

The correct answer is (C).

A differential equation is considered to be ordinary if it has one independent variable. Ordinary differential equations can have as many dependent variables as needed. For example the ordinary differential equations

have two dependent variables y and z, and one independent variable, x.