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Let L(y) = yn+ a1y+a2y where a1, a2are constants, and Let p(r) denote its characterstic polynomial.Then which of the following is/are correct?a)If A.α are constants and P(α)≠0. then Φ(x) = Beax, where B is constant is solution of L(y) = Aeax.b)If Φ(x) and Ψ (x) are solutions of L(y) = b1(x) and L(y) = b2(x), then Φ(x) + Ψ (x) is solutionof L(y) = b1(x) + b2(x).c)If a1= 0 and a2=ω2, whereωis +ve constant, then every solution Φ(x) of L(y) = A cosωx approaches to infinity for large value of x.d)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Let L(y) = yn+ a1y+a2y where a1, a2are constants, and Let p(r) denote its characterstic polynomial.Then which of the following is/are correct?a)If A.α are constants and P(α)≠0. then Φ(x) = Beax, where B is constant is solution of L(y) = Aeax.b)If Φ(x) and Ψ (x) are solutions of L(y) = b1(x) and L(y) = b2(x), then Φ(x) + Ψ (x) is solutionof L(y) = b1(x) + b2(x).c)If a1= 0 and a2=ω2, whereωis +ve constant, then every solution Φ(x) of L(y) = A cosωx approaches to infinity for large value of x.d)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let L(y) = yn+ a1y+a2y where a1, a2are constants, and Let p(r) denote its characterstic polynomial.Then which of the following is/are correct?a)If A.α are constants and P(α)≠0. then Φ(x) = Beax, where B is constant is solution of L(y) = Aeax.b)If Φ(x) and Ψ (x) are solutions of L(y) = b1(x) and L(y) = b2(x), then Φ(x) + Ψ (x) is solutionof L(y) = b1(x) + b2(x).c)If a1= 0 and a2=ω2, whereωis +ve constant, then every solution Φ(x) of L(y) = A cosωx approaches to infinity for large value of x.d)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer?.

Solutions for Let L(y) = yn+ a1y+a2y where a1, a2are constants, and Let p(r) denote its characterstic polynomial.Then which of the following is/are correct?a)If A.α are constants and P(α)≠0. then Φ(x) = Beax, where B is constant is solution of L(y) = Aeax.b)If Φ(x) and Ψ (x) are solutions of L(y) = b1(x) and L(y) = b2(x), then Φ(x) + Ψ (x) is solutionof L(y) = b1(x) + b2(x).c)If a1= 0 and a2=ω2, whereωis +ve constant, then every solution Φ(x) of L(y) = A cosωx approaches to infinity for large value of x.d)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics. Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free.

Here you can find the meaning of Let L(y) = yn+ a1y+a2y where a1, a2are constants, and Let p(r) denote its characterstic polynomial.Then which of the following is/are correct?a)If A.α are constants and P(α)≠0. then Φ(x) = Beax, where B is constant is solution of L(y) = Aeax.b)If Φ(x) and Ψ (x) are solutions of L(y) = b1(x) and L(y) = b2(x), then Φ(x) + Ψ (x) is solutionof L(y) = b1(x) + b2(x).c)If a1= 0 and a2=ω2, whereωis +ve constant, then every solution Φ(x) of L(y) = A cosωx approaches to infinity for large value of x.d)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Let L(y) = yn+ a1y+a2y where a1, a2are constants, and Let p(r) denote its characterstic polynomial.Then which of the following is/are correct?a)If A.α are constants and P(α)≠0. then Φ(x) = Beax, where B is constant is solution of L(y) = Aeax.b)If Φ(x) and Ψ (x) are solutions of L(y) = b1(x) and L(y) = b2(x), then Φ(x) + Ψ (x) is solutionof L(y) = b1(x) + b2(x).c)If a1= 0 and a2=ω2, whereωis +ve constant, then every solution Φ(x) of L(y) = A cosωx approaches to infinity for large value of x.d)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer?, a detailed solution for Let L(y) = yn+ a1y+a2y where a1, a2are constants, and Let p(r) denote its characterstic polynomial.Then which of the following is/are correct?a)If A.α are constants and P(α)≠0. then Φ(x) = Beax, where B is constant is solution of L(y) = Aeax.b)If Φ(x) and Ψ (x) are solutions of L(y) = b1(x) and L(y) = b2(x), then Φ(x) + Ψ (x) is solutionof L(y) = b1(x) + b2(x).c)If a1= 0 and a2=ω2, whereωis +ve constant, then every solution Φ(x) of L(y) = A cosωx approaches to infinity for large value of x.d)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? has been provided alongside types of Let L(y) = yn+ a1y+a2y where a1, a2are constants, and Let p(r) denote its characterstic polynomial.Then which of the following is/are correct?a)If A.α are constants and P(α)≠0. then Φ(x) = Beax, where B is constant is solution of L(y) = Aeax.b)If Φ(x) and Ψ (x) are solutions of L(y) = b1(x) and L(y) = b2(x), then Φ(x) + Ψ (x) is solutionof L(y) = b1(x) + b2(x).c)If a1= 0 and a2=ω2, whereωis +ve constant, then every solution Φ(x) of L(y) = A cosωx approaches to infinity for large value of x.d)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Let L(y) = yn+ a1y+a2y where a1, a2are constants, and Let p(r) denote its characterstic polynomial.Then which of the following is/are correct?a)If A.α are constants and P(α)≠0. then Φ(x) = Beax, where B is constant is solution of L(y) = Aeax.b)If Φ(x) and Ψ (x) are solutions of L(y) = b1(x) and L(y) = b2(x), then Φ(x) + Ψ (x) is solutionof L(y) = b1(x) + b2(x).c)If a1= 0 and a2=ω2, whereωis +ve constant, then every solution Φ(x) of L(y) = A cosωx approaches to infinity for large value of x.d)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? tests, examples and also practice Mathematics tests.