Ordinary Differential Equations Questions and Answers

Answer: C

The value is \(\frac{\sqrt{e}}{2}\).

The characteristic equation \(m^{2} - 2m + 1 = 0\) has the repeated root m = 1, so the general solution is \(y = (A + Bx)e^{x}\).

\[ \begin{aligned} y(0) = 0 &\Rightarrow A = 0 \\ y'(x) = Be^{x}(1 + x) &\Rightarrow y'(0) = B = 1 \\ y &= xe^{x} \end{aligned} \]

At x = 1/2, \(y = \frac{1}{2}e^{1/2} = \frac{\sqrt{e}}{2}\). \(\sqrt{e}\) leaves out the factor x = 1/2, and \(\sqrt{e/2}\) puts the 1/2 under the root. Zero would mean the trivial solution, which the condition y'(0) = 1 rules out.

Answer: C

Remove the radicals by squaring, taking the derivatives as the variables.

\[ \frac{d^2y}{dx^2}\sqrt{\frac{dy}{dx}} = 9 \]

\[ \left(\frac{d^2y}{dx^2}\right)^2 \frac{dy}{dx} = 81 \]

The highest derivative is \(d^2y/dx^2\), so the order is 2, and it appears to the power 2 once the radicals are gone, so the degree is 2. The answer is \(2, 2\), option C. The degree is defined only after the equation is a polynomial in its derivatives, which is why the radicals have to be cleared first.

Answer: C

The general solution is \(Ae^{-2t} + Be^{t}\).

Try \(y = e^{rt}\). The characteristic equation and its roots are:

\[ \begin{aligned} r^{2} + r - 2 &= 0 \\ (r + 2)(r - 1) &= 0 \\ r &= -2,\ 1 \end{aligned} \]

Two distinct real roots give \(y = Ae^{-2t} + Be^{t}\).

The other options use the wrong roots: \(Ae^{-t} + Be^{2t}\) solves \(r^{2} - r - 2 = 0\) (the sign of the middle term flipped), \(Ae^{-2t} + Be^{-t}\) needs roots -2 and -1, and \(Ae^{t} + Be^{2t}\) needs roots 1 and 2. A quick check is that the roots must add to -1 and multiply to -2.

Answer: D

The general solution is \(\sqrt{y} - x + C = 0\).

The equation is separable:

\[ \begin{aligned} \frac{dy}{dx} &= 2\sqrt{y} \\ \frac{dy}{2\sqrt{y}} &= dx \\ \sqrt{y} &= x + c \end{aligned} \]

since the derivative of \(\sqrt{y}\) is \(1/(2\sqrt{y})\). Moving everything to one side gives \(\sqrt{y} - x + C = 0\) with \(C = -c\).

The other forms fail when differentiated: \(y = x - C\) gives \(dy/dx = 1\), and any form with \(\sqrt{x}\) produces a derivative depending on x, which the equation does not have.

Answer: A

The value of \(y\) at \(x = 1\) is 0.

The characteristic equation is \(m^{2} - 4m + 4 = 0\), that is \((m - 2)^{2} = 0\), a repeated root \(m = 2\). So the general solution is \(y = (A + Bx)e^{2x}\).

\[ \begin{aligned} y(0) &= A = 1 \\ y'(x) &= (2A + B + 2Bx)e^{2x} \\ y'(0) &= 2A + B = 1 \Rightarrow B = -1 \\ y(1) &= (1 - 1)e^{2} = 0 \end{aligned} \]

The option \(e^{2}\) is the value of \(e^{2x}\) alone, which satisfies \(y(0) = 1\) but not the slope condition, and neither 1 nor \(e\) fits both conditions.

Answer: B

The solution is \(\frac{1}{3}(1 - e^{-3x})\).

The characteristic equation of \(y'' + 3y' = 0\) is \(m^{2} + 3m = 0\), with roots 0 and -3, so

\[ \begin{aligned} y &= A + Be^{-3x} \\ y(0) = 0 &\Rightarrow A + B = 0 \\ y'(0) = 1 &\Rightarrow -3B = 1,\ B = -\tfrac{1}{3} \end{aligned} \]

Then \(A = \frac{1}{3}\), giving \(y = \frac{1}{3}(1 - e^{-3x})\). Check: \(y' = e^{-3x}\) and \(y'' = -3e^{-3x}\), so \(y'' + 3y' = 0\), with \(y(0) = 0\) and \(y'(0) = 1\).

\(\frac{1}{3}(1 + e^{-3x})\) gives \(y(0) = 2/3\), not 0. \(x(1 - e^{-3x})\) has \(y'(0) = 0\), and \(\frac{1}{3}xe^{-3x/2}\) contains \(x\,e^{-3x/2}\), a form that would need a repeated root at \(-3/2\), which this equation does not have.

Answer: A

The distance is \(\frac{1}{2C_{2}}\ln\left(1 + \frac{C_{2}}{C_{1}}v_{o}^{2}\right)\).

Write the acceleration in terms of distance, \(a = v\,dv/dx\), and separate the variables:

\[ \begin{aligned} v\frac{dv}{dx} &= -(C_{1} + C_{2}v^{2}) \\ dx &= -\frac{v\,dv}{C_{1} + C_{2}v^{2}} \end{aligned} \]

Integrate from \(v_{o}\) down to rest:

\[ \begin{aligned} s &= \frac{1}{2C_{2}}\left[\ln(C_{1} + C_{2}v^{2})\right]_{0}^{v_{o}} \\ &= \frac{1}{2C_{2}}\ln\frac{C_{1} + C_{2}v_{o}^{2}}{C_{1}} \end{aligned} \]

which is the first option. The minus sign version has no physical meaning here, \(\ln(C_{1} + C_{2}v_{o}^{2})\) forgets the lower limit, and \(\ln(1 + C_{2}v_{o}^{2})\) drops the division by \(C_{1}\).

Answer: A

The expression equals 0.

Differentiate \(y = e^{px}\sin qx\):

\[ \begin{aligned} y' &= e^{px}(p\sin qx + q\cos qx) \\ y'' &= e^{px}\left[(p^{2} - q^{2})\sin qx + 2pq\cos qx\right] \end{aligned} \]

Then

\[ \begin{aligned} y'' - 2py' &= e^{px}\left[(-p^{2} - q^{2})\sin qx\right] \\ &= -(p^{2} + q^{2})\,y \end{aligned} \]

so adding \((p^{2} + q^{2})y\) leaves 0. This is expected: the characteristic equation \(m^{2} - 2pm + p^{2} + q^{2} = 0\) has roots \(p \pm iq\), and \(e^{px}\sin qx\) is one of its solutions. The other options would require a forcing term that the equation does not have.

Answer: D

The integrating factor is \(1/(x+1)\).

Divide by \(x + 1\) to put the equation in the standard linear form \(dy/dx + P(x)\,y = Q(x)\):

\[ \begin{aligned} \frac{dy}{dx} - \frac{y}{x+1} &= \frac{\sin x}{x+1} \\ P(x) &= -\frac{1}{x+1} \\ \text{IF} &= e^{\int P\,dx} = e^{-\ln(x+1)} \\ &= \frac{1}{x+1} \end{aligned} \]

Taking \(x + 1\) comes from forgetting the minus sign in \(P(x)\). The options \(x\) and \(1/x\) would belong to an equation with \(x\), not \(x + 1\), as the coefficient of \(dy/dx\).

Answer: A

The solution is \(C = \frac{1}{2}e^{2y} - \frac{1}{3}(e^{3x} + x^{3})\).

Factor out \(e^{-2y}\) and separate the variables:

\[ \begin{aligned} \frac{dy}{dx} &= e^{-2y}(e^{3x} + x^{2}) \\ e^{2y}\,dy &= (e^{3x} + x^{2})\,dx \\ \frac{1}{2}e^{2y} &= \frac{1}{3}e^{3x} + \frac{1}{3}x^{3} + C \end{aligned} \]

Rearranged, \(C = \frac{1}{2}e^{2y} - \frac{1}{3}(e^{3x} + x^{3})\). The options with \(x^{2}\) inside the bracket forget to integrate \(x^{2}\). The last option drops the \(\frac{1}{2}\) from integrating \(e^{2y}\); doubling everything would give \(e^{2y} - \frac{2}{3}(\ldots)\), not \(\frac{1}{3}\).