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The differential equation
For y(x) with the two boundary conditions
For y(x) with the two boundary conditions- a)No solution
- b)Exactly two solutions
- c)Exactly one solution
- d)Infinitely many solutions
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The differential equationFor y(x) with the two boundary conditionsa)No...
Concept:
Given equation is
This is a homogeneous second order differential equation,
So (D2 + 16)y = 0
D2 = m2
⇒ m2 + 16 = 0 ⇒ m = ± 4i = 0 ± 4i
Solution is given as in this case roots are complex, m = α ± i β
y = (C1 cos βx + C2 sin βx) eαx
= (C1 cos 4x + C2 sin 4x) eox = C1 cos 4x + C2 sin 4x
Now y’ = -4C1 sin 4x + 4C2 cos 4x
Applying Boundary condition,
y’ (0) = 1 ⇒ -4C1 sin (0) + 4C2 cos(0) = 1
4C2 = 1 ⇒ C2 = 1/4
Putting another boundary condition.
y′(π/2) = −1
Given equation is
This is a homogeneous second order differential equation,
So (D2 + 16)y = 0
D2 = m2
⇒ m2 + 16 = 0 ⇒ m = ± 4i = 0 ± 4i
Solution is given as in this case roots are complex, m = α ± i β
y = (C1 cos βx + C2 sin βx) eαx
= (C1 cos 4x + C2 sin 4x) eox = C1 cos 4x + C2 sin 4x
Now y’ = -4C1 sin 4x + 4C2 cos 4x
Applying Boundary condition,
y’ (0) = 1 ⇒ -4C1 sin (0) + 4C2 cos(0) = 1
4C2 = 1 ⇒ C2 = 1/4
Putting another boundary condition.
y′(π/2) = −1
So this equation has no solution.
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The differential equationFor y(x) with the two boundary conditionsa)No solutionb)Exactly two solutionsc)Exactly one solutiond)Infinitely many solutionsCorrect answer is option 'A'. Can you explain this answer?
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