First Order Ordinary Differential Equations Questions and Answers

Q: The differential equation by eliminating arbitrary constant \(a\) from :\[ y = a(x - a)^{2} \]is :

The equation is \((y')^{3} - 4xyy' + 8y^{2} = 0\).Write \(p = dy/dx\). From \(y = a(x - a)^{2}\), \(p = 2a(x - a)\). Squaring, \(p^{2} = 4a^{2}(x - a)^{2} = 4ay\), so \(a = p^{2}/(4y)\). Also \(x - a = p/(2a) = 2y/p\), so \(x = p^{2}/(4y) + 2y/p\). Multiply by \(4yp\):\[ \begin{aligned} 4xyp &= p^{3} + 8y^{2} \\ p^{3} - 4xyp + 8y^{2} &= 0 \end{aligned} \]The equation is of degree 3 in \(p\), which rules out the two squared options, and the constant term is \(8y^{2}\), not \(8y\).

Q: The solution of the differential equation\[ y^{1} - 2xy = xy^{2} \]is :

The solution is \(y = \left[-\tfrac{1}{2} + ce^{-x^{2}}\right]^{-1}\).The \(y^{1}\) printed in the equation is \(y'\), so \(y' - 2xy = xy^{2}\) is a Bernoulli equation. Put \(v = 1/y\), so \(v' = -y'/y^{2}\). Dividing by \(y^{2}\):\[ \begin{aligned} -v' - 2xv &= x \\ v' + 2xv &= -x \end{aligned} \]The integrating factor is \(e^{x^{2}}\):\[ \begin{aligned} (ve^{x^{2}})' &= -xe^{x^{2}} \\ ve^{x^{2}} &= -\tfrac{1}{2}e^{x^{2}} + c \\ v &= -\tfrac{1}{2} + ce^{-x^{2}} \end{aligned} \]So \(y = 1/v\). The squared forms do not invert \(v\), and \(e^{+x^{2}}\) comes from the wrong sign of the integrating factor.

Q: Let D be the rectangle :\[ \{(x, y) \in \mathbf{R}^{2} \mid |x| \le 1, |y| \le 1\} \]and \(h\) and \(g\) are functions defined on D given by :\(h(x, y) = xy^{2}\) and\(g(x, y) = y^{2/3}\).Then :

Only \(h\) satisfies the Lipschitz condition on D.The Lipschitz condition here is in \(y\): \(|F(x, y_{1}) - F(x, y_{2})| \le K|y_{1} - y_{2}|\). For \(h = xy^{2}\), \(\partial h/\partial y = 2xy\) and \(|2xy| \le 2\) on D, so by the mean value theorem \(K = 2\) works.For \(g = y^{2/3}\), take \(y_{2} = 0\):\[ \frac{|g(y) - g(0)|}{|y - 0|} = |y|^{-1/3} \to \infty \]as \(y \to 0\), so no constant K works. This is the classic example where \(y' = y^{2/3}\) loses uniqueness. That rules out options naming \(g\) or neither.

Q: Solution of the differential equation\(xdy - ydx = 0,\)represents a :

The solution represents a straight line.Separate the variables:\[ \begin{aligned} \frac{dy}{y} &= \frac{dx}{x} \\ \log|y| &= \log|x| + C \\ y &= cx \end{aligned} \]The general solution is the family of straight lines through the origin. (Equivalently, \(x\,dy - y\,dx = 0\) says \(d(y/x) = 0\).) A hyperbola would come from \(x\,dy + y\,dx = 0\), giving \(xy = c\), and a circle from \(x\,dx + y\,dy = 0\).

Q: The order and degree of the differential equation\[ \begin{aligned} &\left[1 + 3\frac{dy}{dx}\right]^{\frac{2}{3}} \\ &= 4\frac{d^{3}y}{dx^{3}} \end{aligned} \]is :

The order and degree are (3, 3).The highest derivative is \(d^{3}y/dx^{3}\), so the order is 3. To find the degree, first clear the fractional power by cubing both sides:\[ \begin{aligned} &\left(1 + 3\frac{dy}{dx}\right)^{2} \\ &= 64\left(\frac{d^{3}y}{dx^{3}}\right)^{3} \end{aligned} \]The equation is now a polynomial in the derivatives, and the highest derivative appears to the power 3, so the degree is 3.(3, 2) reads the degree from the exponent 2 on the left, which belongs to the first derivative, and the other options miscount the order or degree.

Q: A solution of the differential equation\[ xy' = 2y \]passing through (1, 2), also passes through :

The solution also passes through (2, 8).Separate variables: \(\dfrac{dy}{y} = \dfrac{2\, dx}{x}\), so \(\ln|y| = 2\ln|x| + c\) and \(y = Cx^{2}\). The point (1, 2) gives \(C = 2\), so \(y = 2x^{2}\).At \(x = 2\), \(y = 2 \times 4 = 8\).(2, 4) fits \(y = x^{2}\), the curve through (1, 1), not (1, 2). (2, 16) would need \(y = 2x^{3}\), and (2, 1) lies on none of the solution curves through (1, 2).

Q: The differential equation obtained by eliminating arbitrary constants from the expression \(y = ae^{bx}\) is :

The equation is \(y\dfrac{d^{2}y}{dx^{2}} - \left(\dfrac{dy}{dx}\right)^{2} = 0\).From \(y = ae^{bx}\): \(y' = by\) and \(y'' = b^{2}y\). Eliminating \(b\), \(b = y'/y\), so\[ \begin{aligned} y'' &= \frac{(y')^{2}}{y}, \\ \text{that is,} \quad yy'' - (y')^{2} &= 0 \end{aligned} \]Both constants disappear, as required for a family with two parameters and a second-order equation.The other options contain \(x\) explicitly, but the family \(ae^{bx}\) is unchanged by shifts in \(x\) (a shift only changes \(a\)), so its differential equation cannot involve \(x\).

Q: The initial value problem :\[ y^{1} = f(x, y),\ y(x_{0}) = y_{0} \]where \((x, y)\) belongs to a domain D, has a unique solution :

A unique solution is guaranteed if \(f(x, y)\) is continuous and satisfies a Lipschitz condition in D. (The paper writes \(y^{1}\) for the derivative \(y'\).)This is the Picard-Lindelof theorem: continuity of \(f\) gives existence (Peano's theorem), and the Lipschitz condition \(|f(x, y_{1}) - f(x, y_{2})| \le L|y_{1} - y_{2}|\) makes the successive approximations converge to a single solution and rules out a second one.Continuity alone, even with boundedness, gives existence but not uniqueness. The standard example is \(y' = 3y^{2/3}\), \(y(0) = 0\): \(f\) is continuous and bounded near the origin, yet both \(y = 0\) and \(y = x^{3}\) are solutions, because \(y^{2/3}\) is not Lipschitz at \(y = 0\). And not every function gives a unique solution, so 'for all functions' is false.

Q: For the initial value problem\[ y^{1} = y^{2} + \cos^{2} x,\ y(0) = 0 \]the interval of existence of solution in rectangle\[ \begin{aligned} R : \{(x, y) \mid &0 \le x \le a,\ |y| \le b, \\ &a \gt 1/2,\ b \gt 0\} \end{aligned} \]is :

The interval of existence is [0, 1/2]. (The paper writes \(y^{1}\) for \(y'\).)Picard's theorem on the rectangle \(0 \le x \le a\), \(|y| \le b\) guarantees a solution on \([0, h]\) with \(h = \min(a, b/M)\), where \(M\) bounds \(|f|\) on the rectangle. Here\[ \begin{aligned} M &= \max |y^{2} + \cos^{2} x| = b^{2} + 1 \\ \frac{b}{M} &= \frac{b}{b^{2} + 1} \end{aligned} \]The function \(b/(b^{2} + 1)\) is largest at \(b = 1\), where it equals 1/2. Since \(a \gt 1/2\), the best guaranteed interval is \(h = \min(a, 1/2) = 1/2\), that is [0, 1/2].The intervals [0, 1] and [0, 2] would need \(b/M\) larger than any value it can take, and \([0, 1/\sqrt{2}]\) does not come from the maximisation.

11.

The differential equation by eliminating arbitrary constant \(a\) from :

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

is :

A.\(\left(\dfrac{dy}{dx}\right)^{3} - 4xy\dfrac{dy}{dx} + 8y = 0\)
B.\(\left(\dfrac{dy}{dx}\right)^{3} - 4xy\dfrac{dy}{dx} + 8y^{2} = 0\)
C.\(\left(\dfrac{dy}{dx}\right)^{2} - 4xy\dfrac{dy}{dx} + 8y^{2} = 0\)
D.\(\left(\dfrac{dy}{dx}\right)^{2} - 4xy\dfrac{dy}{dx} + 8y = 0\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The equation is \((y')^{3} - 4xyy' + 8y^{2} = 0\).

Write \(p = dy/dx\). From \(y = a(x - a)^{2}\), \(p = 2a(x - a)\). Squaring, \(p^{2} = 4a^{2}(x - a)^{2} = 4ay\), so \(a = p^{2}/(4y)\). Also \(x - a = p/(2a) = 2y/p\), so \(x = p^{2}/(4y) + 2y/p\). Multiply by \(4yp\):

\[ \begin{aligned} 4xyp &= p^{3} + 8y^{2} \\ p^{3} - 4xyp + 8y^{2} &= 0 \end{aligned} \]

The equation is of degree 3 in \(p\), which rules out the two squared options, and the constant term is \(8y^{2}\), not \(8y\).

Discuss in Forum

12.

The solution of the differential equation

\[ y^{1} - 2xy = xy^{2} \]

is :

A.\(\left[-\dfrac{1}{2} + ce^{-x^{2}}\right]^{2}\)
B.\(\left[-\dfrac{1}{2} + ce^{-x^{2}}\right]^{-1}\)
C.\(\left[-\dfrac{1}{2} + ce^{x^{2}}\right]^{-1}\)
D.\(\left[-\dfrac{1}{2} + ce^{x^{2}}\right]^{2}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The solution is \(y = \left[-\tfrac{1}{2} + ce^{-x^{2}}\right]^{-1}\).

The \(y^{1}\) printed in the equation is \(y'\), so \(y' - 2xy = xy^{2}\) is a Bernoulli equation. Put \(v = 1/y\), so \(v' = -y'/y^{2}\). Dividing by \(y^{2}\):

\[ \begin{aligned} -v' - 2xv &= x \\ v' + 2xv &= -x \end{aligned} \]

The integrating factor is \(e^{x^{2}}\):

\[ \begin{aligned} (ve^{x^{2}})' &= -xe^{x^{2}} \\ ve^{x^{2}} &= -\tfrac{1}{2}e^{x^{2}} + c \\ v &= -\tfrac{1}{2} + ce^{-x^{2}} \end{aligned} \]

So \(y = 1/v\). The squared forms do not invert \(v\), and \(e^{+x^{2}}\) comes from the wrong sign of the integrating factor.

Discuss in Forum

13.

Let D be the rectangle :

\[ \{(x, y) \in \mathbf{R}^{2} \mid |x| \le 1, |y| \le 1\} \]

and \(h\) and \(g\) are functions defined on D given by :
\(h(x, y) = xy^{2}\) and
\(g(x, y) = y^{2/3}\).

Then :

A.Only \(h\) satisfies Lipschitz condition on D
B.Only \(g\) satisfies Lipschitz condition on D
C.Both \(h\) and \(g\) satisfy Lipschitz condition on D
D.Neither \(h\) nor \(g\) satisfies Lipschitz condition on D

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

Only \(h\) satisfies the Lipschitz condition on D.

The Lipschitz condition here is in \(y\): \(|F(x, y_{1}) - F(x, y_{2})| \le K|y_{1} - y_{2}|\). For \(h = xy^{2}\), \(\partial h/\partial y = 2xy\) and \(|2xy| \le 2\) on D, so by the mean value theorem \(K = 2\) works.

For \(g = y^{2/3}\), take \(y_{2} = 0\):

\[ \frac{|g(y) - g(0)|}{|y - 0|} = |y|^{-1/3} \to \infty \]

as \(y \to 0\), so no constant K works. This is the classic example where \(y' = y^{2/3}\) loses uniqueness. That rules out options naming \(g\) or neither.

Discuss in Forum

14.

Solution of the differential equation

\(xdy - ydx = 0,\)

represents a :

A.straight line
B.hyperbola
C.parabola
D.circle

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The solution represents a straight line.

Separate the variables:

\[ \begin{aligned} \frac{dy}{y} &= \frac{dx}{x} \\ \log|y| &= \log|x| + C \\ y &= cx \end{aligned} \]

The general solution is the family of straight lines through the origin. (Equivalently, \(x\,dy - y\,dx = 0\) says \(d(y/x) = 0\).) A hyperbola would come from \(x\,dy + y\,dx = 0\), giving \(xy = c\), and a circle from \(x\,dx + y\,dy = 0\).

Discuss in Forum

15.

The order and degree of the differential equation

\[ \begin{aligned} &\left[1 + 3\frac{dy}{dx}\right]^{\frac{2}{3}} \\ &= 4\frac{d^{3}y}{dx^{3}} \end{aligned} \]

is :

A.(3, 2)
B.(3, 4)
C.(3, 3)
D.(2, 3)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The order and degree are (3, 3).

The highest derivative is \(d^{3}y/dx^{3}\), so the order is 3. To find the degree, first clear the fractional power by cubing both sides:

\[ \begin{aligned} &\left(1 + 3\frac{dy}{dx}\right)^{2} \\ &= 64\left(\frac{d^{3}y}{dx^{3}}\right)^{3} \end{aligned} \]

The equation is now a polynomial in the derivatives, and the highest derivative appears to the power 3, so the degree is 3.

(3, 2) reads the degree from the exponent 2 on the left, which belongs to the first derivative, and the other options miscount the order or degree.

Discuss in Forum

16.

A solution of the differential equation

\[ xy' = 2y \]

passing through (1, 2), also passes through :

A.(2, 1)
B.(2, 4)
C.(2, 8)
D.(2, 16)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The solution also passes through (2, 8).

Separate variables: \(\dfrac{dy}{y} = \dfrac{2\, dx}{x}\), so \(\ln|y| = 2\ln|x| + c\) and \(y = Cx^{2}\). The point (1, 2) gives \(C = 2\), so \(y = 2x^{2}\).

At \(x = 2\), \(y = 2 \times 4 = 8\).

(2, 4) fits \(y = x^{2}\), the curve through (1, 1), not (1, 2). (2, 16) would need \(y = 2x^{3}\), and (2, 1) lies on none of the solution curves through (1, 2).

Discuss in Forum

17.

The differential equation obtained by eliminating arbitrary constants from the expression \(y = ae^{bx}\) is :

A.\(y\dfrac{d^{2}y}{dx^{2}} - \left(\dfrac{dy}{dx}\right)^{2} = 0\)
B.\(y\dfrac{d^{2}y}{dx^{2}} - x\dfrac{dy}{dx} = 0\)
C.\(\dfrac{d^{2}y}{dx^{2}} + xy\left(\dfrac{dy}{dx}\right)^{2} = 0\)
D.\(\dfrac{d^{2}y}{dx^{2}} - xy\dfrac{dy}{dx} = 0\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The equation is \(y\dfrac{d^{2}y}{dx^{2}} - \left(\dfrac{dy}{dx}\right)^{2} = 0\).

From \(y = ae^{bx}\): \(y' = by\) and \(y'' = b^{2}y\). Eliminating \(b\), \(b = y'/y\), so

\[ \begin{aligned} y'' &= \frac{(y')^{2}}{y}, \\ \text{that is,} \quad yy'' - (y')^{2} &= 0 \end{aligned} \]

Both constants disappear, as required for a family with two parameters and a second-order equation.

The other options contain \(x\) explicitly, but the family \(ae^{bx}\) is unchanged by shifts in \(x\) (a shift only changes \(a\)), so its differential equation cannot involve \(x\).

Discuss in Forum

18.

The initial value problem :

\[ y^{1} = f(x, y),\ y(x_{0}) = y_{0} \]

where \((x, y)\) belongs to a domain D, has a unique solution :

A.If \(f(x, y)\) is continuous in D
B.If \(f(x, y)\) is continuous and bounded in D
C.If \(f(x, y)\) is continuous and satisfies Lipschitz condition for all points in D
D.For all functions \(f(x, y)\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

A unique solution is guaranteed if \(f(x, y)\) is continuous and satisfies a Lipschitz condition in D. (The paper writes \(y^{1}\) for the derivative \(y'\).)

This is the Picard-Lindelof theorem: continuity of \(f\) gives existence (Peano's theorem), and the Lipschitz condition \(|f(x, y_{1}) - f(x, y_{2})| \le L|y_{1} - y_{2}|\) makes the successive approximations converge to a single solution and rules out a second one.

Continuity alone, even with boundedness, gives existence but not uniqueness. The standard example is \(y' = 3y^{2/3}\), \(y(0) = 0\): \(f\) is continuous and bounded near the origin, yet both \(y = 0\) and \(y = x^{3}\) are solutions, because \(y^{2/3}\) is not Lipschitz at \(y = 0\). And not every function gives a unique solution, so 'for all functions' is false.

Discuss in Forum

19.

The differential equation of family of circles of radius \(r\) whose center lie on \(x\)-axis is :

A.\(y\left(\dfrac{dy}{dx}\right) + y^{2} = r^{2}\)
B.\(y\left[\left(\dfrac{dy}{dx}\right)^{2} + 1\right] = r^{2}\)
C.\(y^{2}\left[\dfrac{dy}{dx} + 1\right] = r^{2}\)
D.\(y^{2}\left[\left(\dfrac{dy}{dx}\right)^{2} + 1\right] = r^{2}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The differential equation is \(y^{2}\left[\left(\frac{dy}{dx}\right)^{2} + 1\right] = r^{2}\).

A circle of radius \(r\) with centre \((h, 0)\) on the \(x\)-axis is \((x - h)^{2} + y^{2} = r^{2}\), with one arbitrary constant \(h\). Differentiate to eliminate it:

\[ \begin{aligned} 2(x - h) + 2y\,y' &= 0 \\ x - h &= -y\,y' \\ y^{2} y'^{2} + y^{2} &= r^{2} \\ y^{2}\left(y'^{2} + 1\right) &= r^{2} \end{aligned} \]

Option A, \(y y' + y^{2} = r^{2}\), forgets to square \(y y'\). Option B, \(y[(y')^{2} + 1] = r^{2}\), has only one power of \(y\) where the substitution produces \(y^{2}\), and option C, \(y^{2}[y' + 1]\), forgets to square the derivative.

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20.

For the initial value problem

\[ y^{1} = y^{2} + \cos^{2} x,\ y(0) = 0 \]

the interval of existence of solution in rectangle

\[ \begin{aligned} R : \{(x, y) \mid &0 \le x \le a,\ |y| \le b, \\ &a \gt 1/2,\ b \gt 0\} \end{aligned} \]

is :

A.[0, 1]
B.\([0, 1/\sqrt{2}]\)
C.[0, 1/2]
D.[0, 2]

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The interval of existence is [0, 1/2]. (The paper writes \(y^{1}\) for \(y'\).)

Picard's theorem on the rectangle \(0 \le x \le a\), \(|y| \le b\) guarantees a solution on \([0, h]\) with \(h = \min(a, b/M)\), where \(M\) bounds \(|f|\) on the rectangle. Here

\[ \begin{aligned} M &= \max |y^{2} + \cos^{2} x| = b^{2} + 1 \\ \frac{b}{M} &= \frac{b}{b^{2} + 1} \end{aligned} \]

The function \(b/(b^{2} + 1)\) is largest at \(b = 1\), where it equals 1/2. Since \(a \gt 1/2\), the best guaranteed interval is \(h = \min(a, 1/2) = 1/2\), that is [0, 1/2].

The intervals [0, 1] and [0, 2] would need \(b/M\) larger than any value it can take, and \([0, 1/\sqrt{2}]\) does not come from the maximisation.

First Order Ordinary Differential Equations Questions and Answers

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