First Order Ordinary Differential Equations Questions and Answers

Q: The initial value problem :\[ \frac{dy}{dx} = \sqrt{|y|},\quad y(0) = 0, \]

The problem has more than one solution.\(y \equiv 0\) is a solution. So is\[ y(x) = \begin{cases} x^{2}/4, & x \ge 0 \\ 0, & x \lt 0 \end{cases} \]since for \(x \ge 0\), \(y' = x/2 = \sqrt{x^{2}/4} = \sqrt{|y|}\), and \(y(0) = 0\). In fact infinitely many solutions branch off the zero solution at any point \(c \ge 0\).Uniqueness fails because \(\sqrt{|y|}\) is not Lipschitz in y near \(y = 0\). So 'no solution' and 'unique solution' are wrong, and 'has non-zero solution' understates what happens.

Q: Consider the following statements :(I) \(f(x, y) = xy^{2}\) satisfies Lipschitz condition on the rectangle \(|x| \le 1\), \(|y| \le 1\).(II) \(f(x, y) = xy^{2}\) satisfies Lipschitz condition on the strip \(|x| \le 1\), \(|y| \lt \infty\).Then :

Only (I) is true.A function satisfies a Lipschitz condition in y on a region if \(\partial f/\partial y\) is bounded there. Here\[ \frac{\partial f}{\partial y} = 2xy \]On the rectangle \(|x| \le 1\), \(|y| \le 1\), \(|2xy| \le 2\), so a Lipschitz constant 2 works and (I) is true.On the strip, y is unbounded and so is \(2xy\). For x = 1, \(|f(1, y_{1}) - f(1, y_{2})| = |y_{1} + y_{2}|\,|y_{1} - y_{2}|\), and the factor \(|y_{1} + y_{2}|\) can be made as large as we like, so no Lipschitz constant exists and (II) is false.

Q: The initial value problem\[ y' = 2\sqrt{y},\quad y(0) = a \]has :

The problem has infinitely many solutions if a = 0.With \(y(0) = 0\): \(y \equiv 0\) is a solution, and so is, for any \(c \ge 0\),\[ y(x) = \begin{cases} 0, & x \le c \\ (x - c)^{2}, & x \gt c \end{cases} \]since \(y' = 2(x - c) = 2\sqrt{y}\) for \(x \gt c\). Each c gives a different solution, so there are infinitely many.For \(a \lt 0\), \(\sqrt{y}\) is not real near \(x = 0\), so there is no real solution, not a unique one. For \(a \gt 0\) the right side is smooth near \(y = a\), so a solution exists and is unique locally, which rules out 'no solution if a > 0' and, because a = 0 is included, 'a unique solution if a ≥ 0'.

Q: The differential equation corresponding to the family of curves :\[ y = c(x - c)^{2}, \]where c is an arbitrary constant, is :

The differential equation is \((y')^{3} = 4y(xy' - 2y)\).From \(y = c(x - c)^{2}\), \(y' = 2c(x - c)\). Check the right-hand side of option B:\[ \begin{aligned} xy' - 2y &= 2cx(x - c) - 2c(x - c)^{2} \\ &= 2c(x - c)[x - (x - c)] \\ &= 2c^{2}(x - c) \end{aligned} \]\[ \begin{aligned} 4y(xy' - 2y) &= 4c(x - c)^{2} \cdot 2c^{2}(x - c) \\ &= 8c^{3}(x - c)^{3} = (y')^{3} \end{aligned} \]The constant c is eliminated, as it must be for the family's differential equation. The other options fail the same substitution. For option D, for instance, \((y')^{2} = 4c^{2}(x - c)^{2}\), while \(3y(xy' + 2y) = 6c^{2}(x - c)^{3}(2x - c)\), a polynomial of higher degree in x.

Q: For the differential equation :\[ y' = x \sin y + y \cos x \]defined on rectangle\[ |x| \le 5, \; |y| \le 2, \]which of the following is Lipschitz constant ?

A Lipschitz constant is 6.For \(f(x, y) = x \sin y + y \cos x\), the Lipschitz constant in y on the rectangle is the bound of \(|\partial f/\partial y|\):\[ \begin{aligned} \left|\frac{\partial f}{\partial y}\right| &= |x \cos y + \cos x| \\ &\le |x| + 1 \le 5 + 1 = 6 \end{aligned} \]By the mean value theorem, \(|f(x, y_{1}) - f(x, y_{2})| \le 6|y_{1} - y_{2}|\) on \(|x| \le 5\), \(|y| \le 2\).The smaller values do not work: at \(x = 5\), \(y = 0\), the derivative is \(5 + \cos 5 \approx 5.28\), already larger than 4, 3 or 1.

Q: Consider the differential equation\[ \frac{dy}{dx} - y + y^{2} = 0. \]Then \(\displaystyle\lim_{x \to \infty} y(x)\) equals.

The limit is 1.Rewrite the equation as \(y' = y - y^{2} = y(1 - y)\), the logistic equation. Its equilibria are 0 and 1. For \(0 \lt y \lt 1\), \(y' \gt 0\), and for \(y \gt 1\), \(y' \lt 0\), so every solution that starts with \(y(0) \gt 0\) moves towards 1:\[ y(x) = \frac{1}{1 + Ce^{-x}} \to 1 \]The key takes this stable equilibrium as the answer. The value 0 is the unstable equilibrium, reached only by the zero solution, and a solution starting below 0 runs off to \(-\infty\) rather than to \(-1\); \(+\infty\) is never approached.

Q: The general solution of\[ \frac{dy}{dx} + 2xy = 2e^{-x^{2}} \]is :

The general solution is \(y = (2x + c)e^{-x^{2}}\).The equation is linear with integrating factor \(e^{\int 2x\,dx} = e^{x^{2}}\):\[ \begin{aligned} \frac{d}{dx}\left(y e^{x^{2}}\right) &= 2e^{-x^{2}} e^{x^{2}} = 2 \\ y e^{x^{2}} &= 2x + c \end{aligned} \]Check: \(y' = 2e^{-x^{2}} - 2x(2x + c)e^{-x^{2}}\), so \(y' + 2xy = 2e^{-x^{2}}\).\(y = 2ce^{-x^{2}}\) solves only the homogeneous equation, \(ce^{-x^{2}/2}\) uses the wrong exponent, and \((2x^{2} + c)e^{-x^{2}}\) integrates 2 to \(2x^{2}\) instead of \(2x\).

21.

Let \(\Gamma\) be a curve which passes through (0, 1) and intersects each of the curve of the family \(y = cx^{2}\) orthogonally. Then \(\Gamma\) also passes through :

A.\((0, \sqrt{2})\)
B.(1, 1)
C.\((\sqrt{2}, 0)\)
D.(-1, 1)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The orthogonal trajectory through (0, 1) also passes through \((\sqrt{2}, 0)\).

For \(y = cx^{2}\), eliminating \(c = y/x^{2}\) gives the slope \(y' = 2cx = 2y/x\). Orthogonal curves have slope \(-x/(2y)\):

\[ \begin{aligned} 2y\,dy &= -x\,dx \\ y^{2} &= -\frac{x^{2}}{2} + k \\ x^{2} + 2y^{2} &= C \end{aligned} \]

Through (0, 1): \(C = 2\). So \(\Gamma\) is the ellipse \(x^{2} + 2y^{2} = 2\). Checking the options: \((\sqrt{2}, 0)\) gives \(2 + 0 = 2\), so it lies on \(\Gamma\). \((0, \sqrt{2})\) gives 4, while (1, 1) and (-1, 1) give 3, so none of them is on the curve.

Discuss in Forum

22.

The initial value problem :

\[ \frac{dy}{dx} = \sqrt{|y|},\quad y(0) = 0, \]

A.has no solution
B.has unique solution
C.has non-zero solution
D.has more than one solution

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The problem has more than one solution.

\(y \equiv 0\) is a solution. So is

\[ y(x) = \begin{cases} x^{2}/4, & x \ge 0 \\ 0, & x \lt 0 \end{cases} \]

since for \(x \ge 0\), \(y' = x/2 = \sqrt{x^{2}/4} = \sqrt{|y|}\), and \(y(0) = 0\). In fact infinitely many solutions branch off the zero solution at any point \(c \ge 0\).

Uniqueness fails because \(\sqrt{|y|}\) is not Lipschitz in y near \(y = 0\). So 'no solution' and 'unique solution' are wrong, and 'has non-zero solution' understates what happens.

Discuss in Forum

23.

Consider the following statements :

(I) \(f(x, y) = xy^{2}\) satisfies Lipschitz condition on the rectangle \(|x| \le 1\), \(|y| \le 1\).
(II) \(f(x, y) = xy^{2}\) satisfies Lipschitz condition on the strip \(|x| \le 1\), \(|y| \lt \infty\).

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 - 2024, SET Exam - 2024,

Answer: A

Only (I) is true.

A function satisfies a Lipschitz condition in y on a region if \(\partial f/\partial y\) is bounded there. Here

\[ \frac{\partial f}{\partial y} = 2xy \]

On the rectangle \(|x| \le 1\), \(|y| \le 1\), \(|2xy| \le 2\), so a Lipschitz constant 2 works and (I) is true.

On the strip, y is unbounded and so is \(2xy\). For x = 1, \(|f(1, y_{1}) - f(1, y_{2})| = |y_{1} + y_{2}|\,|y_{1} - y_{2}|\), and the factor \(|y_{1} + y_{2}|\) can be made as large as we like, so no Lipschitz constant exists and (II) is false.

Discuss in Forum

24.

The initial value problem

\[ y' = 2\sqrt{y},\quad y(0) = a \]

has :

A.a unique solution if \(a \lt 0\)
B.no solution if \(a \gt 0\)
C.infinitely many solutions if \(a = 0\)
D.a unique solution if \(a \ge 0\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The problem has infinitely many solutions if a = 0.

With \(y(0) = 0\): \(y \equiv 0\) is a solution, and so is, for any \(c \ge 0\),

\[ y(x) = \begin{cases} 0, & x \le c \\ (x - c)^{2}, & x \gt c \end{cases} \]

since \(y' = 2(x - c) = 2\sqrt{y}\) for \(x \gt c\). Each c gives a different solution, so there are infinitely many.

For \(a \lt 0\), \(\sqrt{y}\) is not real near \(x = 0\), so there is no real solution, not a unique one. For \(a \gt 0\) the right side is smooth near \(y = a\), so a solution exists and is unique locally, which rules out 'no solution if a > 0' and, because a = 0 is included, 'a unique solution if a ≥ 0'.

Discuss in Forum

25.

The differential equation corresponding to the family of curves :

\[ y = c(x - c)^{2}, \]

where c is an arbitrary constant, is :

A.\((y')^{2} = 2y(x^{2}y' + 3y)\)
B.\((y')^{3} = 4y(xy' - 2y)\)
C.\((y')^{4} = 3x(x^{3}y' + 4y)\)
D.\((y')^{2} = 3y(xy' + 2y)\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The differential equation is \((y')^{3} = 4y(xy' - 2y)\).

From \(y = c(x - c)^{2}\), \(y' = 2c(x - c)\). Check the right-hand side of option B:

\[ \begin{aligned} xy' - 2y &= 2cx(x - c) - 2c(x - c)^{2} \\ &= 2c(x - c)[x - (x - c)] \\ &= 2c^{2}(x - c) \end{aligned} \]

\[ \begin{aligned} 4y(xy' - 2y) &= 4c(x - c)^{2} \cdot 2c^{2}(x - c) \\ &= 8c^{3}(x - c)^{3} = (y')^{3} \end{aligned} \]

The constant c is eliminated, as it must be for the family's differential equation. The other options fail the same substitution. For option D, for instance, \((y')^{2} = 4c^{2}(x - c)^{2}\), while \(3y(xy' + 2y) = 6c^{2}(x - c)^{3}(2x - c)\), a polynomial of higher degree in x.

Discuss in Forum

26.

For the differential equation :

\[ y' = x \sin y + y \cos x \]

defined on rectangle

\[ |x| \le 5, \; |y| \le 2, \]

which of the following is Lipschitz constant ?

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

A Lipschitz constant is 6.

For \(f(x, y) = x \sin y + y \cos x\), the Lipschitz constant in y on the rectangle is the bound of \(|\partial f/\partial y|\):

\[ \begin{aligned} \left|\frac{\partial f}{\partial y}\right| &= |x \cos y + \cos x| \\ &\le |x| + 1 \le 5 + 1 = 6 \end{aligned} \]

By the mean value theorem, \(|f(x, y_{1}) - f(x, y_{2})| \le 6|y_{1} - y_{2}|\) on \(|x| \le 5\), \(|y| \le 2\).

The smaller values do not work: at \(x = 5\), \(y = 0\), the derivative is \(5 + \cos 5 \approx 5.28\), already larger than 4, 3 or 1.

Discuss in Forum

27.

Consider the differential equation

\[ \frac{dy}{dx} - y + y^{2} = 0. \]

Then \(\displaystyle\lim_{x \to \infty} y(x)\) equals.

A.0
B.1
C.−1
D.\(\infty\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The limit is 1.

Rewrite the equation as \(y' = y - y^{2} = y(1 - y)\), the logistic equation. Its equilibria are 0 and 1. For \(0 \lt y \lt 1\), \(y' \gt 0\), and for \(y \gt 1\), \(y' \lt 0\), so every solution that starts with \(y(0) \gt 0\) moves towards 1:

\[ y(x) = \frac{1}{1 + Ce^{-x}} \to 1 \]

The key takes this stable equilibrium as the answer. The value 0 is the unstable equilibrium, reached only by the zero solution, and a solution starting below 0 runs off to \(-\infty\) rather than to \(-1\); \(+\infty\) is never approached.

Discuss in Forum

28.

The general solution of

\[ \frac{dy}{dx} + 2xy = 2e^{-x^{2}} \]

is :

A.\(y = (2x + c)e^{-x^{2}}\)
B.\(y = 2ce^{-x^{2}}\)
C.\(y = ce^{\frac{-x^{2}}{2}}\)
D.\(y = (2x^{2} + c)e^{-x^{2}}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The general solution is \(y = (2x + c)e^{-x^{2}}\).

The equation is linear with integrating factor \(e^{\int 2x\,dx} = e^{x^{2}}\):

\[ \begin{aligned} \frac{d}{dx}\left(y e^{x^{2}}\right) &= 2e^{-x^{2}} e^{x^{2}} = 2 \\ y e^{x^{2}} &= 2x + c \end{aligned} \]

Check: \(y' = 2e^{-x^{2}} - 2x(2x + c)e^{-x^{2}}\), so \(y' + 2xy = 2e^{-x^{2}}\).

\(y = 2ce^{-x^{2}}\) solves only the homogeneous equation, \(ce^{-x^{2}/2}\) uses the wrong exponent, and \((2x^{2} + c)e^{-x^{2}}\) integrates 2 to \(2x^{2}\) instead of \(2x\).

First Order Ordinary Differential Equations Questions and Answers

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