Question

The complete solution of the linear differential equation

\(\frac{{{d^2}z}}{{d{t^2}}} + \left( {p + q} \right)\frac{{dz}}{{dt}} + pqz = 0\)

1. c₁ e^-pt + c₂ e^-qt
2. c₁ e^pt - c₂ e^qt
3. c₁ e^-pt + c₂ e^qt
4. c₁ e^pt + c₂ e^qt

Answer 1

Correct Answer - Option 1 : c₁ e^-pt + c₂ e^-qt

For the given differential equation:

\(\frac{{{{\rm{d}}^2}{\rm{Z}}}}{{{\rm{d}}{{\rm{t}}^2}}} + \left( {{\rm{p}} + {\rm{q}}} \right)\frac{{{\rm{dz}}}}{{{\rm{dt}}}} + {\rm{pqz}} = 0\)

\({\rm{D}}_{\rm{z}}^2 + \left( {{\rm{p}} + {\rm{q}}} \right){{\rm{D}}_{\rm{z}}} + {\rm{pqz}} = 0\)

(D² + (p + q)D + pq)z = 0

⇒ f(D)z = 0

Where f(D) = D² + (p + q)D + pq

Consider in terms of M ⇒ f(m) = 0

M² + (p + q)m + pq = 0

\({\rm{m}} = \frac{{ - \left( {{\rm{p}} + {\rm{q}}} \right) \pm \sqrt {{{\left( {{\rm{p}} + {\rm{q}}} \right)}^2} - 4{\rm{pq}}} }}{2}\)

\({\rm{m}} = \frac{{ - \left( {{\rm{p}} + {\rm{q}}} \right) \pm \sqrt {{{\rm{p}}^2} + {{\rm{q}}^2} + 2{\rm{pq}} - 4{\rm{pq}}} }}{2}\)

\({\rm{m}} = \frac{{ - \left( {{\rm{p}} + {\rm{q}}} \right) \pm \sqrt {{{\rm{p}}^2} + {{\rm{q}}^2} - 2{\rm{pq}}} }}{2} = - {\rm{\;p}}\)

\({\rm{m}} = \frac{{ - \left( {{\rm{p}} + {\rm{q}}} \right) \pm \sqrt {{{\left( {{\rm{p}} - {\rm{q}}} \right)}^2}} }}{2} = \frac{{ - \left( {{\rm{p}} + {\rm{q}}} \right) \pm \left( {{\rm{p}} - {\rm{q}}} \right)}}{2}\)

\(\Rightarrow {{\rm{m}}_1} = \frac{{ - {\rm{p}} - {\rm{q}} + {\rm{p}} - {\rm{q}}}}{2}\& {\rm{\;}}{{\rm{m}}_2} = \frac{{ - {\rm{p}} - {\rm{q}} - {\rm{p}} + {\rm{q}}}}{2}\)

M₁ = -q & m₂ = -p

Hence, its solution is

Z = C₁e^-pt + C₂e^-qt

∴ Complete solution of the linear differential equation \(\frac{{{d^2}z}}{{d{t^2}}} + \left( {p + q} \right)\frac{{dz}}{{dt}} + pqz = 0\) is c₁ e^-pt + c₂ e^-qt