Differential Equations Questions and Answers

Answer: B

The order is 2 and the degree is 4.

The order is the highest derivative present: \(\frac{d^{2}y}{dx^{2}}\), so 2. For the degree, the equation must first be free of radicals. Raise both sides to the 20th power, the LCM of the root indices 5 and 4:

\[ \begin{aligned} &\left(1 + \frac{d^{2}y}{dx^{2}}\right)^{4} \\ &= \left(y + \left(\frac{dy}{dx}\right)^{5}\right)^{5} \end{aligned} \]

Now the highest-order derivative, \(\frac{d^{2}y}{dx^{2}}\), appears to the power 4, so the degree is 4.

Degree 5 confuses the power of \(\frac{dy}{dx}\) (or of the right side) with the power of the highest derivative. Order 4 reads the root index as an order.

Answer: E

The general solution is \(y = \frac{\sin x + C}{x^{3}}\).

The left side is an exact derivative by the product rule:

\[ \begin{aligned} \frac{d}{dx}(x^{3}y) &= x^{3}\frac{dy}{dx} + 3x^{2}y = \cos x \\ x^{3}y &= \sin x + C \\ y &= \frac{\sin x + C}{x^{3}} \end{aligned} \]

The constant of integration is divided by \(x^{3}\) too, so \(y = \frac{\sin x}{x^{3}} + C\) is wrong: it does not satisfy the equation unless \(C = 0\). The options with \(x^{2}\) in the denominator come from the wrong integrating factor.

Answer: B

The integrating factor is \(e^{-x/2}\).

Put the equation in the standard linear form \(\frac{dy}{dx} + Py = Q\):

\[ \begin{aligned} 2\frac{dy}{dx} &= y + \cos x \\ \frac{dy}{dx} - \tfrac{1}{2}y &= \tfrac{1}{2}\cos x \end{aligned} \]

Here \(P = -\frac{1}{2}\), so the integrating factor is \(e^{\int P\,dx} = e^{-x/2}\). The option \(e^{x/2}\) misses the minus sign that appears when \(y\) is moved to the left, and \(e^{-2x}\) forgets to divide by 2 first.

Answer: D

\(\alpha\) = 3.

Separate the variables and integrate:

\[ \begin{aligned} (2y - 4)\,dy &= (\alpha x^{2} + 4x - 4)\,dx \\ y^{2} - 4y &= \frac{\alpha}{3}x^{3} + 2x^{2} - 4x + c \end{aligned} \]

Comparing with the given \(y^{2} - 4y = x^{3} + 2x^{2} - 4x + c\) gives \(\frac{\alpha}{3} = 1\), so \(\alpha = 3\). The condition y(1) = 3 fixes only c (it gives c = -2) and does not affect \(\alpha\). Taking \(\alpha = 1\) forgets that \(\int x^{2}dx = \frac{x^{3}}{3}\).

Answer: A

The integrating factor is \(\frac{1}{(20 - x)^{2}}\).

The equation is linear, \(\frac{dy}{dx} + P(x)y = Q\), with \(P = \frac{2}{20 - x}\).

\[ \begin{aligned} \int P\,dx &= -2\ln|20 - x| \\ \text{I.F.} &= e^{-2\ln|20 - x|} = \frac{1}{(20 - x)^{2}} \end{aligned} \]

The minus sign comes from the derivative of 20 − x. Missing it gives \((20 - x)^{2}\), not offered; \(\frac{1}{20 - x}\) drops the factor 2, and the log options stop before exponentiating.

Answer: A

The differential equation is \(\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} - 2y = 0\).

\(y = Ae^{x} + Be^{-2x}\) is the general solution of a linear equation whose auxiliary roots are \(m = 1\) and \(m = -2\). The auxiliary equation is

\[ (m - 1)(m + 2) = m^{2} + m - 2 = 0 \]

so the equation is \(y'' + y' - 2y = 0\). Check: \(y' = Ae^{x} - 2Be^{-2x}\), \(y'' = Ae^{x} + 4Be^{-2x}\), and \(y'' + y' - 2y = 0\).

\(y'' - y' - 2y = 0\) has roots 2 and -1, and the equations with \(\mp 2y'\) and \(+y\) have a repeated root, which would give \((A + Bx)e^{\pm x}\).

Answer: C

The solution is \(x^{2} - y^{2} + 4xy = C\).

With \(M = x + 2y\) and \(N = 2x - y\), \(\frac{\partial M}{\partial y} = 2 = \frac{\partial N}{\partial x}\), so the equation is exact. Find \(F\) with \(F_x = M\), \(F_y = N\):

\[ \begin{aligned} F &= \tfrac{x^{2}}{2} + 2xy - \tfrac{y^{2}}{2} \\ F &= c \Rightarrow x^{2} + 4xy - y^{2} = C \end{aligned} \]

Check: differentiating \(x^{2} + 4xy - y^{2}\) gives \((2x + 4y)dx + (4x - 2y)dy\), twice the given form. The coefficients 6, 3 and -4 on \(xy\) fail this check, and \(2x^{2}\) does not match \(M\).

Answer: C

The order is 3 and the degree is 4: 3 and 4.

The order is that of the highest derivative present, \(\frac{d^{3}y}{dx^{3}}\), so it is 3. The degree is defined only after the equation is free of fractional powers of derivatives. Raise both sides to the power 6:

\[ \begin{aligned} \left(1 + y' + y''\right)^{9} \\ &= \left(x + y + y' + y'''\right)^{4} \end{aligned} \]

The highest-order derivative, \(y'''\), now appears to the power 4, so the degree is 4. Reading the degree off the original fractional exponents gives the wrong pairs, and taking \(y''\) as the highest derivative gives order 2.

Answer: D

The solution is \(2y + 3\log|y - 2| = x + c\).

Separate the variables and split the fraction:

\[ \begin{aligned} \frac{2y - 1}{y - 2}\,dy &= dx \\ \left(2 + \frac{3}{y - 2}\right)dy &= dx \\ 2y + 3\log|y - 2| &= x + c \end{aligned} \]

(\(2y - 1 = 2(y - 2) + 3\).) The other options have the wrong coefficient on \(y\), on the logarithm or on \(x\).

Answer: A

The solution is \((e^{y} + 1)\sin x = C\).

Separate the variables by dividing by \((e^{y} + 1)\sin x\):

\[ \begin{aligned} \frac{\cos x}{\sin x}dx + \frac{e^{y}}{e^{y} + 1}dy &= 0 \\ \log\sin x + \log(e^{y} + 1) &= \log C \\ (e^{y} + 1)\sin x &= C \end{aligned} \]

Check: differentiating \((e^{y} + 1)\sin x = C\) gives \((e^{y} + 1)\cos x\,dx + e^{y}\sin x\,dy = 0\), the original equation. The \(\cos x\) version and the \((e^{y} - 1)\) version do not differentiate back to it.