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Solve \(y'=(1-2t)\,y^2\) with \(y(0)=-\tfrac16\) and determine the maximal interval of existence of the solution.
Solution
The equation is separable, \(y'=g(t)h(y)\) with \(g=1-2t\) and \(h=y^2\) . The only zero of \(h\) is \(y=0\) , and the initial value is negative, so the solution stays in \(y<0\) and we may divide:
\[ -\frac1y=t-t^2+C . \]At \(t=0\) , \(-1/y=6\) , so \(C=6\) and
\[ y(t)=\frac{-1}{6+t-t^2}=\frac{1}{t^2-t-6}=\frac{1}{(t-3)(t+2)}. \]The denominator vanishes at \(t=-2\) and \(t=3\) , and the initial time \(0\) lies between them, so the maximal interval of existence is \((-2,3)\) . On it the solution is negative, and it tends to \(-\infty\) at both ends. Although the right-hand side of the equation is a polynomial in \(t\) and \(y\) , the solution exists only on a bounded interval; this is the blow-up of Chapter 3, and the interval could not have been predicted from the equation without solving.
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Solve \(ty'-2y=t^3e^{t}\) for \(t>0\) with \(y(1)=0\) , and say on what interval the existence theorem for this type of equation guarantees the solution.
Answer
Normal form \(y'-\tfrac2ty=t^2e^{t}\) , integrating factor \(t^{-2}\) : \((y/t^2)'=e^{t}\) , \(y=t^2(e^{t}+C)\) , and \(y(1)=0\) gives \(C=-e\) . So \(y=t^2(e^{t}-e)\) , guaranteed on \((0,\infty)\) , the interval containing \(t_0=1\) on which the coefficients are continuous.
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Find the solution of \(y''-4y'+13y=0\) with \(y(0)=1\) , \(y'(0)=8\) and describe its behaviour as \(t\to\infty\) .
Answer
Roots \(2\pm3i\) ; \(y=e^{2t}(\cos3t+2\sin3t)\) . It oscillates with exponentially growing amplitude \(\sqrt5\,e^{2t}\) .
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A fish population obeys \(y'=0.2\,y\,(1-y/1000)-30\) , with \(y\) in tonnes and time in years. Find the equilibria, state which is stable, describe what happens to a population starting at \(150\) tonnes, and find the largest constant harvest rate for which a positive equilibrium exists.
Answer
Equilibria solve \(y^2-1000y+150000=0\) : \(y=500\pm100\sqrt{10}\) , about \(183.8\) and \(816.2\) . The larger is stable. A population of \(150\) is below the unstable equilibrium and collapses to zero in finite time. The growth term has maximum \(0.2\cdot1000/4=50\) , so a positive equilibrium exists only for harvest rates up to \(50\) tonnes per year.
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Show that \(y'=y^{1/3}\) , \(y(0)=0\) has at least three solutions on \([0,\infty)\) , explain why this does not contradict the existence and uniqueness theorem, and find the solution with \(y(0)=1\) together with the largest interval on which it is the only solution through its points.
Solution
Three solutions. The function \(y=0\) is a solution. For \(t\ge0\) let \(y=(2t/3)^{3/2}\) . Then \(y'=\tfrac32(2t/3)^{1/2}\cdot\tfrac23=(2t/3)^{1/2}=y^{1/3}\) , and \(y(0)=0\) with \(y'(0)=0\) . Its negative, \(-(2t/3)^{3/2}\) , is a third solution, since the cube root is an odd function.
No contradiction. The existence and uniqueness theorem requires \(f(t,y)=y^{1/3}\) to be Lipschitz in \(y\) near the initial point. The quotient \(\lvert y^{1/3}\rvert/\lvert y\rvert=\lvert y\rvert^{-2/3}\) is unbounded near \(y=0\) , so the hypothesis fails at every point of the \(t\) -axis, and the theorem makes no claim there.
The solution through \((0,1)\) . Separating in the region \(y>0\) : \(\tfrac32y^{2/3}=t+C\) , and \(y(0)=1\) gives \(C=\tfrac32\) , so
\[ y=\Bigl(1+\frac{2t}{3}\Bigr)^{3/2},\qquad t>-\frac32 . \]For \(y>0\) the right-hand side has the continuous derivative \(\tfrac13y^{-2/3}\) , so through each point of this curve the solution is unique. As \(t\to-\tfrac32\) the solution tends to \(0\) and reaches the line where uniqueness fails; continued to the left it could remain at \(0\) or not, so \((-\tfrac32,\infty)\) is the largest interval on which it is the only solution through its points.
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Find the steady-state response of \(y''+4y'+5y=10\cos t\) in the form \(R\cos(t-\phi)\) , and explain why every solution approaches it.
Solution
Replace \(10\cos t\) by \(10e^{it}\) . With \(P(r)=r^2+4r+5\) , \(P(i)=-1+4i+5=4+4i\) , and
\[ z_p=\frac{10e^{it}}{4+4i}=\frac{10\,(4-4i)}{32}e^{it}=\frac54(1-i)(\cos t+i\sin t). \]The real part is \(y_p=\tfrac54(\cos t+\sin t)\) . Since \(4+4i=4\sqrt2\,e^{i\pi/4}\) , the amplitude is \(R=10/(4\sqrt2)=\tfrac{5\sqrt2}{4}\approx1.768\) and the lag is \(\phi=\pi/4\) :
\[ y_p=\tfrac{5\sqrt2}{4}\cos\bigl(t-\tfrac\pi4\bigr). \]The characteristic roots are \(-2\pm i\) , with negative real part. Every solution is \(y_p\) plus a combination of \(e^{-2t}\cos t\) and \(e^{-2t}\sin t\) , which tends to zero, so every solution approaches the steady state regardless of initial conditions.
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A system at rest obeys \(y''+2y'+y=\delta(t-2)\) . Find \(y(t)\) , the time at which the displacement is largest, and the largest displacement.
Answer
\((s+1)^2Y=e^{-2s}\) , so \(y=u(t-2)\,(t-2)\,e^{-(t-2)}\) . The function \(\tau e^{-\tau}\) is largest at \(\tau=1\) , so the maximum displacement is \(e^{-1}\approx0.368\) at \(t=3\) .
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Compute \(e^{tA}\) for \(A=\begin{pmatrix}-1&2\\-2&-1\end{pmatrix}\) , classify the origin of \(\mathbf x'=A\mathbf x\) , and state the direction of rotation.
Answer
\(A=-I+2K\) with \(K=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\) , \(K^2=-I\) , so \(e^{tA}=e^{-t}\begin{pmatrix}\cos2t&\sin2t\\-\sin2t&\cos2t\end{pmatrix}\) . Eigenvalues \(-1\pm2i\) : a stable spiral, clockwise since \(a_{21}<0\) .
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For \(x''=-x+x^3\) find the equilibria of the equivalent first-order system, a conserved quantity, and the stability of each equilibrium, noting where linearization alone is inconclusive.
Answer
\(x'=y\) , \(y'=-x+x^3\) ; equilibria \((0,0)\) and \((\pm1,0)\) . \(E=\tfrac12y^2+\tfrac12x^2-\tfrac14x^4\) is conserved. At \((\pm1,0)\) the Jacobian \(\begin{pmatrix}0&1\\2&0\end{pmatrix}\) has eigenvalues \(\pm\sqrt2\) : saddles, unstable. At \((0,0)\) the eigenvalues are \(\pm i\) and linearization is inconclusive; \(E\) has a strict minimum there, so the origin is stable, a centre.
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Find the recurrence relation for power series solutions of \(y''+t^2y=0\) about \(t=0\) and the first two nonzero terms of each of two independent solutions.
Answer
\(a_2=a_3=0\) and \(a_{n+2}=-\dfrac{a_{n-2}}{(n+2)(n+1)}\) for \(n\ge2\) . \(y_1=1-\dfrac{t^4}{12}+\cdots\) , \(y_2=t-\dfrac{t^5}{20}+\cdots\) .
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Solve \((2xy+y^2)\,dx+(x^2+2xy)\,dy=0\) and give the solution curve through \((1,1)\) .
Answer
\(M_y=2x+2y=N_x\) , exact, with potential \(x^2y+xy^2\) . Through \((1,1)\) : \(x^2y+xy^2=2\) .
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For \(y'=y^3-y\) draw the phase line, and decide for which initial values \(y(0)\) the solution exists for all \(t\ge0\) , giving a reason in each case.
Answer
\(f(y)=y(y-1)(y+1)\) : equilibria \(-1,0,1\) , with \(0\) stable and \(\pm1\) unstable. If \(\lvert y(0)\rvert\le1\) the solution is trapped between equilibria and exists for all \(t\ge0\) . If \(\lvert y(0)\rvert>1\) then \(\lvert y\rvert\) increases and \(\lvert y'\rvert\) is eventually at least a positive multiple of \(\lvert y\rvert^3\) , so the solution blows up in finite time.
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Find the general solution of \(y''-2y'+y=e^{t}/t\) for \(t>0\) .
Solution
The characteristic polynomial is \((r-1)^2\) , so \(y_1=e^{t}\) , \(y_2=te^{t}\) form a fundamental set, with Wronskian \(e^{t}(1+t)e^{t}-e^{t}\cdot te^{t}=e^{2t}\) . The forcing \(g=e^{t}/t\) is not of the kind handled by undetermined coefficients, so use variation of parameters:
\[ u_1'=-\frac{y_2g}{W}=-\frac{te^{t}\cdot e^{t}/t}{e^{2t}}=-1,\qquad u_2'=\frac{y_1g}{W}=\frac{e^{t}\cdot e^{t}/t}{e^{2t}}=\frac1t . \]So \(u_1=-t\) , \(u_2=\ln t\) , and \(y_p=-te^{t}+te^{t}\ln t\) . The first term is a multiple of \(y_2\) and can be absorbed into the complementary solution. The general solution for \(t>0\) is
\[ y=e^{t}\bigl(c_1+c_2t+t\ln t\bigr). \]Check with \(y=e^{t}v\) , for which the equation reduces to \(v''=1/t\) as in Example 5.3: \(v=t\ln t\) has \(v'=\ln t+1\) and \(v''=1/t\) .
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For which \(\omega\) does \(y''+9y=\cos\omega t\) have unbounded solutions? For \(y''+0.2\,y'+9y=\cos\omega t\) find the forcing frequency that maximizes the steady-state amplitude, and that maximum amplitude.
Answer
Undamped: only \(\omega=3\) , the natural frequency. Damped: \(\omega_r=\sqrt{9-0.02}\approx2.997\) , and the maximum gain is \(1/(0.2\sqrt{9-0.01})\approx1.668\) .
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Show in two ways that the origin is an asymptotically stable equilibrium of \(x'=-x+xy\) , \(y'=-2y-x^2\) , once by linearization and once with \(V=x^2+y^2\) , and say what the second method proves that the first does not.
Answer
The Jacobian at the origin is diagonal with entries \(-1\) and \(-2\) , so Theorem 8.3(a) gives asymptotic stability. With \(V=x^2+y^2\) : \(\dot V=2x(-x+xy)+2y(-2y-x^2)=-2x^2-4y^2<0\) away from the origin, so Theorem 8.6(b) applies on the whole plane. The second method shows that every solution converges, whereas linearization gives only an unspecified neighbourhood.
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Solve \(\mathbf x'=\begin{pmatrix}2&1\\0&2\end{pmatrix}\mathbf x+\begin{pmatrix}0\\e^{2t}\end{pmatrix}\) with \(\mathbf x(0)=\mathbf 0\) .
Solution
The matrix is \(A=2I+N\) with \(N=\begin{pmatrix}0&1\\0&0\end{pmatrix}\) , \(N^2=0\) , so \(e^{tA}=e^{2t}\begin{pmatrix}1&t\\0&1\end{pmatrix}\) . With zero initial data, variation of constants gives
\[ \mathbf x(t)=\int_0^te^{(t-s)A}\begin{pmatrix}0\\e^{2s}\end{pmatrix}ds=\int_0^te^{2(t-s)}e^{2s}\begin{pmatrix}t-s\\1\end{pmatrix}ds=e^{2t}\int_0^t\begin{pmatrix}t-s\\1\end{pmatrix}ds=e^{2t}\begin{pmatrix}t^2/2\\t\end{pmatrix}. \]Check: \(x_2=te^{2t}\) satisfies \(x_2'=2x_2+e^{2t}\) , and \(x_1=\tfrac12t^2e^{2t}\) satisfies \(x_1'=te^{2t}+t^2e^{2t}=2x_1+x_2\) . The forcing \(e^{2t}\) resonates with the eigenvalue \(2\) , giving the factor \(t\) in \(x_2\) ; and because the matrix is defective, \(x_2\) in turn drives \(x_1\) resonantly, giving \(t^2\) .