Differential Equations Questions and Answers

Answer: C

The solution is \(y = 1 - 2x^{2}\).

Separate and integrate, then fix the constant from \(y = 1\) at \(x = 0\):

\[ \begin{aligned} \frac{dy}{dx} &= -4x \\ y &= \int -4x\,dx = -2x^{2} + C \\ y(0) = 1 &\Rightarrow C = 1 \\ y &= 1 - 2x^{2} \end{aligned} \]

Check: differentiating gives \(-4x\), and \(-4x + 4x = 0\). The option \(y = -4x^{2} + 1\) forgets to halve on integrating, so its derivative is \(-8x\). \(y = 4x\) gives \(y = 0\) at \(x = 0\), and \(\log(x)\) is not even defined at \(x = 0\).

Answer: B

The value is \(\sqrt{\frac{7 + \sqrt{53}}{2}}\).

Factor both sides and separate the variables:

\[ \begin{aligned} \frac{dy}{dx} &= \frac{y^{3}(e^{5x} + 1)}{e^{x}(1 + y^{4})} \\ \left(\frac{1}{y^{3}} + y\right)dy &= \left(e^{4x} + e^{-x}\right)dx \end{aligned} \]

Integrating both sides:

\[ \begin{aligned} -\frac{1}{2y^{2}} + \frac{y^{2}}{2} \\ = \frac{e^{4x}}{4} - e^{-x} + C \end{aligned} \]

At \(x = 0\), \(y^{2} = \frac{1}{2}\): the left side is \(-1 + \frac{1}{4} = -\frac{3}{4}\) and the right side is \(\frac{1}{4} - 1 + C\), so \(C = 0\). At \(x = \log_{e} 2\), \(e^{4x} = 16\) and \(e^{-x} = \frac{1}{2}\). Put \(u = y^{2}\):

\[ \begin{aligned} \frac{u}{2} - \frac{1}{2u} &= 4 - \frac{1}{2} = \frac{7}{2} \\ u^{2} - 7u - 1 &= 0 \\ u &= \frac{7 + \sqrt{53}}{2} \end{aligned} \]

The positive root is taken because \(u = y^{2} \gt 0\), and \(y \gt 0\) by the stated range, so \(y = \sqrt{u}\).

\(\frac{7 + \sqrt{53}}{2}\) is \(y^{2}\), not \(y\): it forgets the square root. The pair built on \(\sqrt{35}\) does not satisfy \(u^{2} - 7u - 1 = 0\).

Answer: D

\(y(x) = \) \(e^{2}e^{x/2}\).

The equation is \(\frac{dy}{dx} = \frac{y}{2}\). Separating variables:

\[ \begin{aligned} \frac{dy}{y} &= \frac{dx}{2} \\ \ln y &= \frac{x}{2} + c \\ y &= Ce^{x/2} \end{aligned} \]

\(y(0) = C = e^{2}\), so \(y = e^{2}e^{x/2}\). \(e^{x/2}\) has the right shape but gives \(y(0) = 1\); \(e^{2x}\) solves \(y' = 2y\), which reverses the given relation; the options with \(+ 2\) do not satisfy the equation at all.

Answer: E

The integrating factor is cosec x.

Since \(\frac{1}{2}\sin 2x = \sin x\cos x\), dividing by \(\sin x\,dx\) gives the linear form:

\[ \begin{aligned} \frac{dy}{dx} - (\cot x)\,y &= \cos x \\ \text{I.F.} &= e^{-\int\cot x\,dx} = e^{-\ln\sin x} = \frac{1}{\sin x} \end{aligned} \]

So the integrating factor is \(\csc x\). \(\sin x\) is what you get by forgetting the minus sign on the \(y\) term, and sec x, tan x and cos x come from integrating \(\cot x\) wrongly.

Answer: C

The solution is \(x - y + xy = C\).

Separate the variables:

\[ \begin{aligned} \frac{dx}{1 - x} &= \frac{dy}{1 + y} \\ -\log|1 - x| &= \log|1 + y| + c \\ (1 - x)(1 + y) &= k \end{aligned} \]

Expanding, \(1 + y - x - xy = k\), that is \(x - y + xy = C\). Check: differentiating \(x - y + xy = C\) gives \((1 + y)dx + (x - 1)dy = 0\), the given equation. The other options give different equations on differentiation.

Answer: A

The integrating factor is \(x^{2} + 1\).

Put the equation in linear form:

\[ \frac{dy}{dx} + \frac{2x}{1 + x^{2}}y = \frac{1}{1 + x^{2}} \]

The integrating factor is \(e^{\int P\,dx} = e^{\log(1 + x^{2})} = 1 + x^{2}\). \(\log_{e}(x^{2} + 1)\) stops before taking the exponential, which is the usual slip; the other options do not come from \(\int\frac{2x}{1 + x^{2}}dx\).

Multiplying through by \(1 + x^{2}\) gives \(\frac{d}{dx}[(1 + x^{2})y] = 1\), so the solution is \((1 + x^{2})y = x + C\).

Answer: A

The answer is \(y = \sqrt{16 + \sqrt{25 + \sqrt{x}}} + C\).

Differentiate the candidate with the chain rule, from the outside in.

\[ \begin{aligned} \frac{d}{dx}\sqrt{16 + \sqrt{25 + \sqrt{x}}} &= \frac{1}{2\sqrt{16 + \sqrt{25 + \sqrt{x}}}}\cdot\frac{d}{dx}\sqrt{25 + \sqrt{x}} \\ \frac{d}{dx}\sqrt{25 + \sqrt{x}} &= \frac{1}{2\sqrt{25 + \sqrt{x}}}\cdot\frac{1}{2\sqrt{x}} \end{aligned} \]

The three factors of \(\frac{1}{2}\) give the 8 in the denominator, and the three roots match exactly, so this is the given \(\frac{dy}{dx}\). Adding x or \(x^{2}\), or multiplying by them, would add terms that are not there.

Answer: C

The order is 3.

The order equals the number of independent arbitrary constants. \(c_{1}\) and \(c_{2}\) appear only as the sum \(c_{1} + c_{2}\), which is a single constant A. So the family is \(y = A\sin(x + c_{3}) - c_{4}e^{x}\).

That leaves three independent constants, A, \(c_{3}\) and \(c_{4}\), so eliminating them needs three differentiations and gives a third-order equation. Counting the four letters gives 4, which overlooks that \(c_{1}\) and \(c_{2}\) cannot be separated.

Answer: E

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

Recognise the two exact differentials: \(xdy + ydx = d(xy)\) and \(\frac{ydx - xdy}{y^{2}} = d\left(\frac{x}{y}\right)\). Divide the equation by \(y^{2}\):

\[ \begin{aligned} \frac{ydx - xdy}{y^{2}} &= xdy + ydx \\ d\left(\frac{x}{y}\right) &= d(xy) \end{aligned} \]

Integrating, \(\frac{x}{y} = xy + C\). \(\frac{y}{x}\) has the quotient upside down, since \(d\left(\frac{y}{x}\right) = \frac{xdy - ydx}{x^{2}}\), with the opposite sign and \(x^{2}\) underneath.

Answer: C

The solution is \(ye^{x} = x + 1\).

The equation is linear with P = 1, so the integrating factor is \(e^{\int 1\,dx} = e^{x}\):

\[ \begin{aligned} \frac{d}{dx}(ye^{x}) &= e^{-x} \cdot e^{x} = 1 \\ ye^{x} &= x + C \end{aligned} \]

At x = 0, y = 1 gives C = 1, so \(ye^{x} = x + 1\). \(ye^{x} = x - 1\) has the wrong constant (it gives y = -1 at x = 0), and the options with \(e^{-x}\) use the reciprocal of the integrating factor.