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Let x, x + ex and 1 + x + ex be solutions of a linear second order ordinary differential equation with constant coefficients. If y(x) is the solution of the same equation satisfying y(0) = 3 and y'(0) = 4, then y(1) is equal to
- a)3e + 1
- b)2e + 3
- c)3e + 2
- d)e + 1
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Let x, x + ex and 1 + x + ex be solutions of a linear second order ord...
Here x, x + ex are two linear independent
solution. so general solution can be written as
y = c1x + c2 (x + ex)
solution. so general solution can be written as
y = c1x + c2 (x + ex)
= (c1+ c2) x + c2ex
y′= (c1+ c2) + c2ex
.
y(0) = c2 = 3
y′ (0) = 4 = c1+ 2c2
.
y(0) = c2 = 3
y′ (0) = 4 = c1+ 2c2
⇒ c1 = 4 – 6 = –2
so, y(x) (–2 + 3) x + 3ex = 3ex+ x
so, y(x) (–2 + 3) x + 3ex = 3ex+ x
y(1) = 3e + 1
Most Upvoted Answer
Let x, x + ex and 1 + x + ex be solutions of a linear second order ord...
Solution:
Given that x, x ex and 1 x ex are solutions of a linear second order ordinary differential equation with constant coefficients.
Let the differential equation be y'' + py' + qy = 0.
Then, the characteristic equation is r² + pr + q = 0.
Given that x and x ex are solutions of the differential equation.
Substituting y = x in the differential equation, we get
y'' + py' + qy = x'' + px' + qx = 0
=> p + p + q = 0
=> 2p + q = 0
Substituting y = x ex in the differential equation, we get
y'' + py' + qy = (ex)'' + p(ex)' + q(ex) = 0
=> qex = ex
=> q = 1
Substituting y = 1 x in the differential equation, we get
y'' + py' + qy = (1 x)'' + p(1 x)' + q(1 x) = 0
=> 2 + (p 2)x + px' + qx = 0
=> 2 + (p 2)x + px' + x = 0
=> (p 1)x' + (x + 2) = 0
Integrating both sides with respect to x, we get
ln(x + 2) px x = C
where C is a constant of integration.
Using the initial conditions y(0) = 3 and y'(0) = 4, we get
C = ln 5 and p = 2.
Hence, the solution of the differential equation satisfying the initial conditions is
y = (ln(x + 2) 2x + 1)e x
Therefore, y(1) = [(ln 3 2 + 1)e] = 3e 1.
Hence, the correct answer is option 'A'.
Given that x, x ex and 1 x ex are solutions of a linear second order ordinary differential equation with constant coefficients.
Let the differential equation be y'' + py' + qy = 0.
Then, the characteristic equation is r² + pr + q = 0.
Given that x and x ex are solutions of the differential equation.
Substituting y = x in the differential equation, we get
y'' + py' + qy = x'' + px' + qx = 0
=> p + p + q = 0
=> 2p + q = 0
Substituting y = x ex in the differential equation, we get
y'' + py' + qy = (ex)'' + p(ex)' + q(ex) = 0
=> qex = ex
=> q = 1
Substituting y = 1 x in the differential equation, we get
y'' + py' + qy = (1 x)'' + p(1 x)' + q(1 x) = 0
=> 2 + (p 2)x + px' + qx = 0
=> 2 + (p 2)x + px' + x = 0
=> (p 1)x' + (x + 2) = 0
Integrating both sides with respect to x, we get
ln(x + 2) px x = C
where C is a constant of integration.
Using the initial conditions y(0) = 3 and y'(0) = 4, we get
C = ln 5 and p = 2.
Hence, the solution of the differential equation satisfying the initial conditions is
y = (ln(x + 2) 2x + 1)e x
Therefore, y(1) = [(ln 3 2 + 1)e] = 3e 1.
Hence, the correct answer is option 'A'.
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Let x, x + ex and 1 + x + ex be solutions of a linear second order ordinary differential equation with constant coefficients. If y(x) is the solution of the same equation satisfying y(0) = 3 and y(0) = 4, then y(1) is equal toa)3e + 1b)2e + 3c)3e + 2d)e + 1Correct answer is option 'A'. Can you explain this answer?
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Let x, x + ex and 1 + x + ex be solutions of a linear second order ordinary differential equation with constant coefficients. If y(x) is the solution of the same equation satisfying y(0) = 3 and y(0) = 4, then y(1) is equal toa)3e + 1b)2e + 3c)3e + 2d)e + 1Correct answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let x, x + ex and 1 + x + ex be solutions of a linear second order ordinary differential equation with constant coefficients. If y(x) is the solution of the same equation satisfying y(0) = 3 and y(0) = 4, then y(1) is equal toa)3e + 1b)2e + 3c)3e + 2d)e + 1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let x, x + ex and 1 + x + ex be solutions of a linear second order ordinary differential equation with constant coefficients. If y(x) is the solution of the same equation satisfying y(0) = 3 and y(0) = 4, then y(1) is equal toa)3e + 1b)2e + 3c)3e + 2d)e + 1Correct answer is option 'A'. Can you explain this answer?.
Let x, x + ex and 1 + x + ex be solutions of a linear second order ordinary differential equation with constant coefficients. If y(x) is the solution of the same equation satisfying y(0) = 3 and y(0) = 4, then y(1) is equal toa)3e + 1b)2e + 3c)3e + 2d)e + 1Correct answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let x, x + ex and 1 + x + ex be solutions of a linear second order ordinary differential equation with constant coefficients. If y(x) is the solution of the same equation satisfying y(0) = 3 and y(0) = 4, then y(1) is equal toa)3e + 1b)2e + 3c)3e + 2d)e + 1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let x, x + ex and 1 + x + ex be solutions of a linear second order ordinary differential equation with constant coefficients. If y(x) is the solution of the same equation satisfying y(0) = 3 and y(0) = 4, then y(1) is equal toa)3e + 1b)2e + 3c)3e + 2d)e + 1Correct answer is option 'A'. Can you explain this answer?.
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