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For the Ordinary Differential Equation
with initial conditions x(0) = 0 and dx/dt(0) = 10, the solution is
with initial conditions x(0) = 0 and dx/dt(0) = 10, the solution is- a)-5e2t + 6e3t
- b)5e2t = + 6e3t
- c)-10e2t + 10e3t
- d)10e2t + 10e+3t
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
For the Ordinary Differential Equationwith initial conditions x(0) = 0...
Above given equation is a linear differential equation of order = 2. Such equations are solved using CF + PI method.
Let D = dx/dt
Let D = dx/dt
⇒ D2x – 5Dx + 6x = 0
⇒ (D2 – 5D + 6)x = 0
Thus auxiliary equation (obtained by replacing D with m) is m2 – 5D + 6 = 0.
Roots of above obtained auxiliary equation are-
m2 – 5D + 6 = 0
⇒ (m – 2) (m - 3) = 0
⇒ m1 = 2 or m2 = 3
When both roots of auxiliary equation are real and distinct
Thus general solution of above D.E is
Applying boundary conditions, x(0) = 0
⇒ O = C1 + C2
⇒ C1 = -C2 ---(i)
Using initial condition dx/dt(0) = 10
⇒ 10 = 2C1 + 3C2 ---(ii)
Solving equation (i) & (ii) simultaneously,
C1 = -10 and C2 = 10
Thus, x = -10e2t + 10e3t
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For the Ordinary Differential Equationwith initial conditions x(0) = 0 anddx/dt(0) = 10,the solution isa)-5e2t+ 6e3tb)5e2t= + 6e3tc)-10e2t+ 10e3td)10e2t+ 10e+3tCorrect answer is option 'C'. Can you explain this answer?
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