Solution.

(a)

From the second equation \(x = y'\), so \(x' = y''\). Substituting into the first, \[y'' = 2y - y' \implies y''+y'-2y = 0 .\] The auxiliary equation \(m^2+m-2 = (m+2)(m-1) = 0\) gives \(y = Ae^{t} + Be^{-2t}\), and then \(x = y' = Ae^{t} - 2Be^{-2t}\).

The conditions \(y(0)=0\) and \(x(0)=1\) give \(A+B = 0\) and \(A-2B = 1\), so \(A = \dfrac 13\) and \(B = -\dfrac 13\). Hence \[x = \frac {e^{t}+2e^{-2t}}{3},\qquad y = \frac {e^{t}-e^{-2t}}{3}.\]

(b)

The first equation is uncoupled: \(x = Ae^{3t}\). Then \[y'-y = Ae^{3t}\] is linear with integrating factor \(e^{-t}\), giving \(\big (ye^{-t}\big )' = Ae^{2t}\) and \(y = \dfrac {A}{2}e^{3t} + Be^{t}\). The equation for \(z\) is identical in form, so \[z = \frac {A}{2}e^{3t} + Ce^{t}.\]

(c)

From the first equation \(y = \dfrac {4x-x'}{2}\), so \(y' = \dfrac {4x'-x''}{2}\). Substituting into the second and simplifying, \[x'' - 5x' + 6x = 0 ,\] with roots \(m = 2, 3\), so \(x = Ae^{2t}+Be^{3t}\) and \(y = \dfrac {4x-x'}{2} = Ae^{2t} + \dfrac {B}{2}e^{3t}\).

The conditions \(x(0)=0\), \(y(0)=1\) give \(A+B = 0\) and \(A+\dfrac {B}{2} = 1\), so \(A = 2\) and \(B = -2\): \[x = 2e^{2t}-2e^{3t},\qquad y = 2e^{2t}-e^{3t}.\]

Solution.

(a)

Transforming, with \(X = \mathcal {L}\{x\}\) and \(Y = \mathcal {L}\{y\}\), \[sX-1 = 5X+Y,\qquad sY+3 = X+5Y ,\] that is \[(s-5)X - Y = 1,\qquad -X + (s-5)Y = -3 .\] Solving, \[X = \frac {(s-5)-3}{(s-5)^2-1} = \frac {s-8}{(s-6)(s-4)},\qquad Y = \frac {-3(s-5)+1}{(s-6)(s-4)} = \frac {-3s+16}{(s-6)(s-4)}.\] Partial fractions give \[X = \frac {-1}{s-6} + \frac {2}{s-4},\qquad Y = \frac {-1}{s-6} + \frac {-2}{s-4},\] so \[x = 2e^{4t}-e^{6t},\qquad y = -2e^{4t}-e^{6t}.\] Substituting back into the original equations confirms both.

(b)

Transforming with zero initial conditions, \[2sX + sY - 2X = \frac 1s,\qquad sX+sY-3X-3Y = \frac 2s .\] That is \[(2s-2)X + sY = \frac 1s,\qquad (s-3)X + (s-3)Y = \frac 2s .\] The second gives \(X+Y = \dfrac {2}{s(s-3)}\), so \(Y = \dfrac {2}{s(s-3)} - X\). Substituting into the first, \[(2s-2)X + s\Big [\frac {2}{s(s-3)}-X\Big ] = \frac 1s \implies (s-2)X = \frac 1s - \frac {2}{s-3},\] so \[X = \frac {(s-3)-2s}{s(s-3)(s-2)} = \frac {-(s+3)}{s(s-2)(s-3)} .\] Partial fractions give \[X = -\frac {1}{2s} + \frac {5}{2(s-2)} - \frac {2}{s-3},\] hence \[x(t) = -\frac 12 + \frac 52e^{2t} - 2e^{3t},\] and \(y\) follows from \(y = \displaystyle \mathcal {L}^{-1}\Big \{\frac {2}{s(s-3)}\Big \} - x\), that is \[y(t) = \frac 23\big (e^{3t}-1\big ) - x(t) = -\frac 16 - \frac 52e^{2t} + \frac 83e^{3t}.\] Both satisfy the equations and vanish at \(t=0\), which is the check worth making: an arithmetic slip in the partial fractions shows up at once as a non-zero initial value.

Solution. Writing the equations as \[y'+y-x = -t,\qquad x'+2x-2y = t ,\] the first gives \(x = y'+y+t\), so \(x' = y''+y'+1\). Substituting into the second, \[y''+y'+1 + 2\big (y'+y+t\big ) - 2y = t ,\] \[y'' + 3y' = -t - 1 .\] The complementary function is \(y_c = A + Be^{-3t}\), from \(m^2+3m = 0\). For a particular integral, the forcing term is a polynomial of degree \(1\) but the constant already appears in \(y_c\), so try \(y_p = at^2+bt\): \[2a + 3\big (2at+b\big ) = -t-1 \implies 6a = -1,\quad 2a+3b = -1 ,\] giving \(a = -\dfrac 16\) and \(b = -\dfrac {2}{9}\). Hence \[y = A + Be^{-3t} - \frac {t^2}{6} - \frac {2t}{9},\] and \[x = y'+y+t = A - 2Be^{-3t} - \frac {t^2}{6} + \frac {4t}{9} - \frac {2}{9}.\]

Solution.

(a)

With \(y = \sum a_nx^n\), \[\sum (n+2)(n+1)a_{n+2}x^n + \sum na_nx^n + \sum a_nx^n = 0 ,\] so \((n+2)(n+1)a_{n+2} + (n+1)a_n = 0\) and \[a_{n+2} = \frac {-a_n}{n+2}.\] Taking \(a_0 = 1\), \(a_1 = 0\): \(a_2 = -\dfrac 12\), \(a_4 = \dfrac 18\), so \[y_1 = 1 - \frac {x^2}{2} + \frac {x^4}{8} - \cdots = \sum _{k}\frac {(-1)^kx^{2k}}{2^kk!} = e^{-x^2/2},\] which may be checked directly. Taking \(a_0=0\), \(a_1=1\) gives \[y_2 = x - \frac {x^3}{3} + \frac {x^5}{15} - \cdots \]

(b)

\[\sum (n+2)(n+1)a_{n+2}x^n = \sum a_nx^{n+1} \implies a_{n+2} = \frac {a_{n-1}}{(n+2)(n+1)},\] with \(a_2 = 0\) from the constant term. The recurrence links indices three apart, so the solutions come in the chains driven by \(a_0\) and \(a_1\), and every third coefficient vanishes: \[y = a_0\Big (1+\frac {x^3}{6}+\frac {x^6}{180}+\cdots \Big ) + a_1\Big (x+\frac {x^4}{12}+\frac {x^7}{504}+\cdots \Big ).\] These are the Airy functions.

(c)

\[(n+2)(n+1)a_{n+2} + a_{n-2} = 0 \implies a_{n+2} = \frac {-a_{n-2}}{(n+2)(n+1)},\] linking indices four apart. With \(y(0)=1\) and \(y'(0)=0\) we have \(a_0=1\), \(a_1=0\), and \(a_2 = a_3 = 0\) from the low-order terms. Then \[a_4 = \frac {-1}{12},\qquad a_8 = \frac {-a_4}{56} = \frac {1}{672},\] so \[y = 1 - \frac {x^4}{12} + \frac {x^8}{672} - \cdots \]

Solution. Write each in the form \(y''+P(x)y'+Q(x)y = 0\). The point \(x=0\) is ordinary if \(P\) and \(Q\) are analytic there, regular singular if \(xP\) and \(x^2Q\) are, and irregular otherwise.

(a)

\(P = 0\) and \(Q = \dfrac {1}{x\big (1+x^2\big )}\). Then \(x^2Q = \dfrac {x}{1+x^2}\) is analytic at \(0\), so \(x=0\) is regular singular.

(b)

\(P = \dfrac {1}{x^2}\) and \(Q = \dfrac {1}{x^2}\). Here \(xP = \dfrac 1x\) is not analytic at \(0\), so \(x=0\) is irregular singular.

(c)

\(P = 0\) and \(Q = \dfrac {-1}{\sin x}\). Since \(\sin x = x - \dfrac {x^3}{6}+\cdots \) has a simple zero at the origin, \(x^2Q = \dfrac {-x^2}{\sin x}\) is analytic there, so \(x=0\) is regular singular.

(d)

\(P = 0\) and \(Q = -\tan x\), both analytic at \(0\) since \(\cos 0 = 1\neq 0\). So \(x=0\) is an ordinary point.

Solution. Try \(y = \displaystyle \sum _{n=0}^{\infty }a_nx^{n+r}\) with \(a_0\neq 0\). Then \[x^2y'' = \sum a_n(n+r)(n+r-1)x^{n+r},\] so the equation becomes \[\sum a_n\Big [(n+r)(n+r-1) - 1\Big ]x^{n+r} = 0 .\] The lowest power gives the indicial equation \[r(r-1)-1 = 0,\qquad \text {that is}\qquad r^2-r-1 = 0 ,\] with roots \[r = \frac {1\pm \sqrt 5}{2}.\] For \(n\geq 1\) the coefficient of \(x^{n+r}\) forces \(a_n\big [(n+r)(n+r-1)-1\big ] = 0\), and the bracket is non-zero for those roots, so every \(a_n\) with \(n\geq 1\) vanishes. Each root therefore contributes a single term, and since the roots differ by \(\sqrt 5\), which is not an integer, the two solutions are independent: \[y = Ax^{(1+\sqrt 5)/2} + Bx^{(1-\sqrt 5)/2}.\] This is in fact an Euler–Cauchy equation, and the same answer follows from trying \(y = x^m\) directly — a useful check that the Frobenius machinery has been applied correctly.