Higher Order Linear Differential Equations Questions and Answers

Q: Consider the following two statements :(I) The set \(\{x, |x|\}\) is linearly independent on \((-\infty, \infty)\)(II) The Wronskian \(W(x, |x|) = 0\) on \((-\infty, \infty)\)Then :

The keyed answer is both (I) and (II) are true.(I): if \(c_{1}x + c_{2}|x| = 0\) for all \(x\), then \(x = 1\) gives \(c_{1} + c_{2} = 0\) and \(x = -1\) gives \(-c_{1} + c_{2} = 0\), so \(c_{1} = c_{2} = 0\). The set is linearly independent.(II): for \(x \gt 0\), \(|x| = x\) and \(W = x \cdot 1 - x \cdot 1 = 0\); for \(x \lt 0\), \(|x| = -x\) and \(W = x(-1) - (-x)(1) = 0\). At \(x = 0\), \(|x|\) is not differentiable, so the Wronskian vanishes wherever it is defined, which is the textbook reading of (II).This is the standard example showing that a zero Wronskian does not imply dependence when the functions are not solutions of one linear ODE.

Q: If \(\phi_{1}\) and \(\phi_{2}\) are two linearly independent solutions of \(y'' + ay = 0\), where \(a\) is a constant, on an interval I. Then which of the following statements about the Wronskin \(W(\phi_{1} \phi_{2})(x)\) is true ?

The Wronskian is a constant function.For \(y'' + p(x)y' + q(x)y = 0\), Abel's formula gives \(W(x) = W(x_{0})\exp\left(-\int_{x_{0}}^{x} p(t)\, dt\right)\). Here there is no \(y'\) term, so \(p = 0\) and \(W\) is constant. Directly: \(W' = \phi_{1}\phi_{2}'' - \phi_{1}''\phi_{2}\), which is \(\phi_{1}(-a\phi_{2}) - (-a\phi_{1})\phi_{2} = 0\).Because \(\phi_{1}, \phi_{2}\) are linearly independent, the constant is non-zero, so D (identically zero) is false, and a non-zero constant is neither increasing nor decreasing.

Q: Let \(f(x) = x^{2}\) and \(g(x) = x^{2} \log x\) on \(|x| \gt 0\). Then :

The correct statement is that \(f\) and \(g\) can be solutions of an ordinary differential equation.They are linearly independent, since \(g/f = \log x\) is not constant, which rules out 'linearly dependent'. Their Wronskian is\[ \begin{aligned} W &= x^{2}(2x\log x + x) - 2x \cdot x^{2}\log x \\ &= x^{3} \end{aligned} \]so it is not \(x^{2}\). Since \(W \neq 0\) for \(x \neq 0\), they form a fundamental pair for a second-order linear equation. Indeed \(x^{2}\) and \(x^{2}\log x\) are the solutions of the Euler equation \(x^{2}y'' - 3xy' + 4y = 0\), whose indicial equation \(m^{2} - 4m + 4 = 0\) has the repeated root 2. So 'cannot be solutions of an ordinary differential equation' is false.

Q: Consider the following statements for any solution \(\phi\) of the differential equation with constant coefficient :\[ y'' + a_{1}y' + a_{2}y = 0, \]where \(a_{1}, a_{2}\) are constants.(I) \(\phi \to 0\) as \(x \to \infty\), if the real parts of the roots of the characteristic polynomial are negative.(II) \(\phi \to 0\) as \(x \to \infty\), if the real parts of the roots of the characteristic polynomial are positive.Then :

Only I is true.Every solution is a combination of terms \(e^{rx}\), \(x e^{rx}\) or \(e^{\alpha x}\cos\beta x\), \(e^{\alpha x}\sin\beta x\), where \(r\) or \(\alpha \pm i\beta\) are the characteristic roots. If every root has negative real part, each term carries a decaying exponential and \(\phi \to 0\).If the real parts are positive, the terms grow: for \(y'' - 3y' + 2y = 0\) the roots are 1 and 2 and \(\phi = e^{x}\) tends to infinity. So II is false (only the zero solution would tend to 0).

11.

Consider the following two statements :

(I) The set \(\{x, |x|\}\) is linearly independent on \((-\infty, \infty)\)
(II) The Wronskian \(W(x, |x|) = 0\) on \((-\infty, \infty)\)

Then :

A.Both (I) and (II) are true
B.Neither (I) nor (II) is true
C.Only (I) is true
D.Only (II) is true

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The keyed answer is both (I) and (II) are true.

(I): if \(c_{1}x + c_{2}|x| = 0\) for all \(x\), then \(x = 1\) gives \(c_{1} + c_{2} = 0\) and \(x = -1\) gives \(-c_{1} + c_{2} = 0\), so \(c_{1} = c_{2} = 0\). The set is linearly independent.

(II): for \(x \gt 0\), \(|x| = x\) and \(W = x \cdot 1 - x \cdot 1 = 0\); for \(x \lt 0\), \(|x| = -x\) and \(W = x(-1) - (-x)(1) = 0\). At \(x = 0\), \(|x|\) is not differentiable, so the Wronskian vanishes wherever it is defined, which is the textbook reading of (II).

This is the standard example showing that a zero Wronskian does not imply dependence when the functions are not solutions of one linear ODE.

Discuss in Forum

12.

If \(\phi_{1}\) and \(\phi_{2}\) are two linearly independent solutions of \(y'' + ay = 0\), where \(a\) is a constant, on an interval I. Then which of the following statements about the Wronskin \(W(\phi_{1} \phi_{2})(x)\) is true ?

A.\(W(\phi_{1}, \phi_{2})(x)\) is a constant function
B.\(W(\phi_{1} \phi_{2})(x)\) is a monotonically increasing function
C.\(W(\phi_{1} \phi_{2})(x)\) is a monotonically decreasing function
D.\(W(\phi_{1} \phi_{2})(x) = 0\) for all \(x \in I\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The Wronskian is a constant function.

For \(y'' + p(x)y' + q(x)y = 0\), Abel's formula gives \(W(x) = W(x_{0})\exp\left(-\int_{x_{0}}^{x} p(t)\, dt\right)\). Here there is no \(y'\) term, so \(p = 0\) and \(W\) is constant. Directly: \(W' = \phi_{1}\phi_{2}'' - \phi_{1}''\phi_{2}\), which is \(\phi_{1}(-a\phi_{2}) - (-a\phi_{1})\phi_{2} = 0\).

Because \(\phi_{1}, \phi_{2}\) are linearly independent, the constant is non-zero, so D (identically zero) is false, and a non-zero constant is neither increasing nor decreasing.

Discuss in Forum

13.

Let \(f(x) = x^{2}\) and \(g(x) = x^{2} \log x\) on \(|x| \gt 0\). Then :

A.\(f, g\) are linearly dependent
B.\(f, g\) are linearly independent but cannot be solutions of an ordinary differential equation
C.\(f, g\) can be solutions of an ordinary differential equation
D.Wronskian of \(f\) and \(g\) is \(x^{2}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The correct statement is that \(f\) and \(g\) can be solutions of an ordinary differential equation.

They are linearly independent, since \(g/f = \log x\) is not constant, which rules out 'linearly dependent'. Their Wronskian is

\[ \begin{aligned} W &= x^{2}(2x\log x + x) - 2x \cdot x^{2}\log x \\ &= x^{3} \end{aligned} \]

so it is not \(x^{2}\). Since \(W \neq 0\) for \(x \neq 0\), they form a fundamental pair for a second-order linear equation. Indeed \(x^{2}\) and \(x^{2}\log x\) are the solutions of the Euler equation \(x^{2}y'' - 3xy' + 4y = 0\), whose indicial equation \(m^{2} - 4m + 4 = 0\) has the repeated root 2. So 'cannot be solutions of an ordinary differential equation' is false.

Discuss in Forum

14.

Consider the following statements for any solution \(\phi\) of the differential equation with constant coefficient :

\[ y'' + a_{1}y' + a_{2}y = 0, \]

where \(a_{1}, a_{2}\) are constants.

(I) \(\phi \to 0\) as \(x \to \infty\), if the real parts of the roots of the characteristic polynomial are negative.
(II) \(\phi \to 0\) as \(x \to \infty\), if the real parts of the roots of the characteristic polynomial are positive.

Then :

A.Only I is true
B.Only II is true
C.Both I and II are true
D.Both I and II are false

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

Only I is true.

Every solution is a combination of terms \(e^{rx}\), \(x e^{rx}\) or \(e^{\alpha x}\cos\beta x\), \(e^{\alpha x}\sin\beta x\), where \(r\) or \(\alpha \pm i\beta\) are the characteristic roots. If every root has negative real part, each term carries a decaying exponential and \(\phi \to 0\).

If the real parts are positive, the terms grow: for \(y'' - 3y' + 2y = 0\) the roots are 1 and 2 and \(\phi = e^{x}\) tends to infinity. So II is false (only the zero solution would tend to 0).

Higher Order Linear Differential Equations Questions and Answers

Topics

Other Subjects

Tags: Higher Order Linear Differential Equations Questions and Answers

About Us

Higher Order Linear Differential Equations Questions and Answers

Topics

Other Subjects

Tags: Higher Order Linear Differential Equations Questions and Answers

About Us

Stay informed about all important Dates & Notifications