First Order Ordinary Differential Equations Questions and Answers

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 \(8y\).

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.

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.

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 \(x\,dx + y\,dy = 0\).

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.

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).

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 \(x\).