Ordinary Differential Equations Questions and Answers

Answer: A

The general solution is \(e^{3x}(a\cos 4x + b\sin 4x)\).

The characteristic equation of \(y'' - 6y' + 25y = 0\) is

\[ \begin{aligned} m^{2} - 6m + 25 &= 0 \\ m &= \frac{6 \pm \sqrt{36 - 100}}{2} = 3 \pm 4i \end{aligned} \]

Complex roots \(\alpha \pm i\beta\) give \(y = e^{\alpha x}(a\cos\beta x + b\sin\beta x)\), here \(\alpha = 3\), \(\beta = 4\).

The option with \(e^{4x}\) and \(\cos 3x\) swaps the real and imaginary parts, and the options with \(e^{3ix}\) or \(e^{4ix}\) put an imaginary exponent where the real part of the root belongs, which is not a real solution form.

Answer: D

It solves \(\frac{d^{2}y}{dx^{2}} - m^{2}y = 0\).

Differentiate twice:

\[ \begin{aligned} y' &= ame^{mx} - bme^{-mx} \\ y'' &= am^{2}e^{mx} + bm^{2}e^{-mx} = m^{2}y \end{aligned} \]

So \(y'' - m^{2}y = 0\). The two arbitrary constants a and b show that the equation is of second order, which rules out both first-order options: \(y' - my = 0\) is solved by \(ae^{mx}\) alone and \(y' + my = 0\) by \(be^{-mx}\) alone. \(y'' + m^{2}y = 0\) has the oscillating solutions cos mx and sin mx, not exponentials.

Answer: B

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

The auxiliary equation of \(y'' + 2y' - 3y = 0\) is

\[ m^{2} + 2m - 3 = (m - 1)(m + 3) = 0 \]

so \(m = 1\) or \(-3\) and \(y = Ae^{x} + Be^{-3x}\). Apply the conditions:

\[ \begin{aligned} y(0) &= A + B = 0 \\ y'(0) &= A - 3B = 8 \\ \Rightarrow A &= 2,\ B = -2 \end{aligned} \]

The options with \(e^{3x}\) take the wrong root (+3, from \(m^{2} - 2m - 3\)). \(2(e^{x} + e^{-3x})\) has the right roots but gives \(y(0) = 4\), not 0.

Answer: A

It is a second order, nonhomogeneous, and linear ordinary differential equation.

The highest derivative is \(y''\), so the order is two. The unknown \(y\) and its derivatives appear only to the first power and are multiplied only by functions of \(x\), never by each other, so the equation is linear. The right-hand side \(r(x)\) does not involve \(y\); when it is not identically zero the equation is nonhomogeneous.

A homogeneous equation would have \(r(x) = 0\), so the two 'homogeneous' options are wrong, and nothing in the equation is nonlinear. 'First order' ignores the \(y''\) term.

Answer: A

The general solution is \(y = -\frac{\cos 2x}{4} + C_{1}x + C_{2}\).

Integrate twice:

\[ \begin{aligned} \frac{dy}{dx} &= \int \cos 2x\, dx = \frac{\sin 2x}{2} + C_{1} \\ y &= \int \left( \frac{\sin 2x}{2} + C_{1} \right) dx \\ &= -\frac{\cos 2x}{4} + C_{1}x + C_{2} \end{aligned} \]

A second-order equation needs two arbitrary constants, which rules out both options with only \(C_{1}\); \(\frac{\sin 2x}{2} + C_{1}\) is in fact only the first integral. The option with \(+\frac{\cos 2x}{4}\) has the wrong sign: its second derivative is \(-\cos 2x\).

Answer: A

The problem has no solution.

The general solution is \(y = A\cos 4x + B\sin 4x\), so \(y' = -4A\sin 4x + 4B\cos 4x\).

\[ \begin{aligned} y'(0) &= 4B = 1 \Rightarrow B = \tfrac{1}{4} \\ y'(\pi/2) &= -4A\sin 2\pi + 4B\cos 2\pi = 4B = 1 \end{aligned} \]

The second condition demands -1, but every solution gives y'(\(\pi\)/2) = 1, so the two conditions contradict each other.

If the second condition had been +1, A would be free and there would be infinitely many solutions; a linear problem like this can never have exactly two.

Answer: C

The value is 0.

The equation is separable:

\[ \begin{aligned} y^{3}\,dy &= -x^{3}\,dx \\ \frac{y^{4}}{4} &= -\frac{x^{4}}{4} + C \\ y^{4} &= -x^{4} + 4C \end{aligned} \]

The condition y(0) = 1 gives 4C = 1, so \(y^{4} = 1 - x^{4}\). At x = -1, \(y^{4} = 1 - 1 = 0\), so y(-1) = 0.

The solution is the curve \(x^{4} + y^{4} = 1\), which meets the axis y = 0 at x = ±1, so 1, -1 and -2 cannot be values of y there.

Answer: C

The solution is \(\tfrac{1}{2}x^{2}u^{2} + 2u = \text{constant}\).

Cross-multiply to put the equation in the form \(M\,dx + N\,du = 0\):

\[ xu^{2}\,dx + (2 + x^{2}u)\,du = 0 \]

It is exact, since \(\partial M/\partial u = 2xu = \partial N/\partial x\). Find the potential \(\psi(x, u)\):

\[ \begin{aligned} \psi &= \int xu^{2}\,dx = \tfrac{1}{2}x^{2}u^{2} + h(u) \\ \psi_{u} &= x^{2}u + h'(u) = 2 + x^{2}u \\ h(u) &= 2u \end{aligned} \]

So \(\tfrac{1}{2}x^{2}u^{2} + 2u = C\). Differentiating it back gives \(xu^{2}\,dx + (x^{2}u + 2)\,du = 0\), the original equation.

The others fail the same check. \(u^{2} + 2x^{2}\) gives \(du/dx = -2x/u\); \(xu^{2} + u\) gives \(du/dx = -u^{2}/(2xu + 1)\); and \(\tfrac{1}{2}ux^{2} + 2x\) gives \(du/dx = -(2ux + 4)/x^{2}\). None matches \(-xu^{2}/(2 + x^{2}u)\).

Answer: A

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

This is a Cauchy-Euler equation. Try \(y = x^{m}\):

\[ \begin{aligned} m(m-1)+m-4 &= 0 \\ m^{2} &= 4,\ m = \pm 2 \\ y &= c_{1}x^{2}+c_{2}x^{-2} \end{aligned} \]

The condition y(0) = 0 needs \(c_{2}=0\), since \(x^{-2}\) is unbounded at the origin, and y(1) = 1 then gives \(c_{1}=1\).

The three sine options satisfy both boundary values but not the equation: substituting \(\sin(\pi x/2)\), for example, leaves terms in \(x^{2}\sin\) and \(x\cos\) that do not cancel.

Answer: C

The general solution is \(c_{1}e^{x} + c_{2}xe^{x} + c_{3}\cos(x) + c_{4}\sin(x)\).

The characteristic equation factors neatly:

\[ \begin{aligned} r^{4} - 2r^{3} + 2r^{2} - 2r + 1 &= 0 \\ (r^{2} - 2r + 1)(r^{2} + 1) &= 0 \\ (r - 1)^{2}(r^{2} + 1) &= 0 \end{aligned} \]

The double root r = 1 gives \(e^{x}\) and \(xe^{x}\), and the roots \(\pm i\) give \(\cos x\) and \(\sin x\).

The cosh/sinh option treats \(\pm 1\) as roots; the options with \(e^{-x}\) or \(xe^{-x}\) also assume a root at -1, which does not exist. Expanding the factors confirms the coefficients -2, 2, -2, 1.