Practice / Review Problems — MATH 2410

Solution:

Since the decay rate is proportional to the amount present, the model is \(\frac{dx}{dt}=kx\), whose general solution is \(x(t)=Ce^{kt}\). Since \(x(0)=100\), we have \(C=100\), so

\[ x(t) = 100e^{kt}. \]

The half-life tells us that \(x(5) = 50\):

\[ 50 = 100e^{5k} \quad\Longrightarrow\quad e^{5k}=\frac{1}{2} \quad\Longrightarrow\quad k=\frac{1}{5}\ln\!\left(\frac{1}{2}\right) = -\frac{\ln 2}{5}. \]

Rather than compute \(k\) numerically, it is cleaner to note that \(20=4\cdot 5\), so \(e^{20k}=(e^{5k})^4\). Using \(e^{5k}=\frac12\),

\[ x(20) = 100e^{20k} = 100\left(e^{5k}\right)^4 = 100\left(\frac12\right)^4 = \frac{100}{16}. \]

So

\[ \boxed{x(20) = 6.25 \text{ mg}.} \]

Solution:

Since the growth rate is proportional to the population, \(\frac{dP}{dt}=kP\), so \(P(t) = Ce^{kt}\) for constants \(C\) and \(k\) that we still need to find. Here \(C\) is not given directly (the first data point is not at \(t=0\)), so we use both pieces of data together. From the two observations,

\[ P(3) = Ce^{3k} = 400, \qquad P(10) = Ce^{10k} = 2000. \]

Dividing the second equation by the first eliminates \(C\) entirely, since it appears in both:

\[ \frac{Ce^{10k}}{Ce^{3k}} = \frac{2000}{400} \quad\Longrightarrow\quad e^{7k} = 5. \]

Solving for \(k\):

\[ k = \frac{\ln 5}{7}. \]

Now substitute this back into \(Ce^{3k}=400\) to solve for \(C\), which is the initial population \(P(0)\):

\[ C = 400e^{-3k} = 400e^{-3\ln(5)/7} = 400\cdot 5^{-3/7}. \]

Numerically, \(5^{-3/7}\approx 0.5017\), so

\[ \boxed{P(0) = C \approx 201 \text{ bacteria}.} \]

Solution:

By Newton's Law of Cooling, \(\frac{dT}{dt}=k(T-T_a)\) with ambient temperature \(T_a=0\), so the equation simplifies to \(\frac{dT}{dt}=kT\), and

\[ T(t) = Ce^{kt}. \]

Using the two data points:

\[ T(1)=Ce^{k}=20, \qquad T(5)=Ce^{5k}=5. \]

Dividing the second equation by the first eliminates \(C\):

\[ \frac{Ce^{5k}}{Ce^{k}}=\frac{5}{20} \quad\Longrightarrow\quad e^{4k}=\frac{1}{4} \quad\Longrightarrow\quad k = \frac{1}{4}\ln\!\left(\frac{1}{4}\right) = -\frac{\ln 4}{4}. \]

Substitute back into \(Ce^k=20\) to solve for \(C\), the room's initial temperature:

\[ C = 20e^{-k} = 20e^{(\ln 4)/4} = 20\cdot 4^{1/4}. \]

Since \(4^{1/4}=\sqrt{2}\approx 1.4142\),

\[ \boxed{T(0) = C = 20\sqrt{2} \approx 28.3^\circ\text{C}.} \]

Solution:

This problem is harder than the previous cooling problems because the ambient temperature \(T_m\) (the freezer's temperature) is itself unknown, so we have three unknowns (\(T_m\), \(k\), and the constant of integration) but only two equations from the given readings — the trick is to use the initial reading to eliminate the integration constant, leaving two equations in \(T_m\) and \(k\), and then eliminate \(k\) algebraically.

By Newton's Law of Cooling, \(T(t) = T_m + Ce^{kt}\). Since \(T(0)=72\), we get \(C=72-T_m\), so

\[ T(t) = T_m + (72-T_m)e^{kt}. \]

Using the two readings:

\[ T(2) = T_m+(72-T_m)e^{2k} = 40 \;\Longrightarrow\; e^{2k}=\frac{40-T_m}{72-T_m}, \] \[ T(4) = T_m+(72-T_m)e^{4k} = 24 \;\Longrightarrow\; e^{4k}=\frac{24-T_m}{72-T_m}. \]

Since \(e^{4k}=(e^{2k})^2\), we can eliminate \(k\) entirely by squaring the first relation and setting it equal to the second:

\[ \frac{24-T_m}{72-T_m} = \left(\frac{40-T_m}{72-T_m}\right)^2 \quad\Longrightarrow\quad (24-T_m)(72-T_m) = (40-T_m)^2. \]

Expanding both sides:

\[ 1728-96T_m+T_m^2 = 1600-80T_m+T_m^2. \]

The \(T_m^2\) terms cancel, leaving a linear equation:

\[ 1728-96T_m = 1600-80T_m \quad\Longrightarrow\quad 128 = 16T_m. \]

So

\[ \boxed{T_m = 8^\circ\text{F}.} \]

As a check: with \(T_m=8\), \(e^{2k}=\frac{40-8}{72-8}=\frac{32}{64}=\frac12\), so \(e^{4k}=\frac14\), and \(T_m+(72-T_m)e^{4k}=8+64\left(\frac14\right)=8+16=24\), which matches the second reading exactly.

Solution:

Linearity is about how the unknown function (\(x\) or \(T\)) appears in the equation, not about how the equation is used. In \(\frac{dx}{dt}=kx\), the unknown \(x\) appears only to the first power on both sides, so it is linear. In \(\frac{dT}{dt}=k(T-T_m)\), expanding the right-hand side gives \(kT-kT_m\); since \(T_m\) is a constant, this is still just \(T\) to the first power plus a constant term, so the equation is linear in \(T\) as well — it fits the general linear form \(T'-kT=-kT_m\).

If instead \(\frac{dx}{dt}=-kx^2\), the unknown \(x\) is squared on the right-hand side, so the equation is nonlinear. This does not make the equation unsolvable — it is still separable, since we can write \(\frac{dx}{x^2}=-k\,dt\) — but it does mean the clean exponential solution \(x=Ce^{kt}\) that works for every problem in this section no longer applies. Separating and integrating \(\frac{dx}{x^2}=-k\,dt\) instead produces \(-\frac1x=-kt+C\), i.e. \(x=\frac{1}{kt+C}\), a rational function of \(t\) rather than an exponential one. This is exactly the kind of nonlinear behavior that Section 3.2 explores in depth (the logistic equation is built from a nonlinear rate like this one).

Solution:

Comparing to the standard logistic form \(\frac{dP}{dt}=P(a-bP)\), we read off \(a=5\) and \(b=0.001\). The carrying capacity is where the per-capita growth rate \(a-bP\) vanishes:

\[ 5-0.001P=0 \quad\Longrightarrow\quad \boxed{K = \frac{a}{b} = \frac{5}{0.001} = 5000.} \]

When \(P\) is very small compared to \(K=5000\), the term \(0.001P\) is negligible compared to \(5\), so the equation behaves approximately like

\[ \frac{dP}{dt} \approx 5P, \]

i.e. the population grows approximately exponentially at rate \(5\) whenever it is far below its carrying capacity — the logistic equation only "feels" the carrying capacity once \(P\) becomes comparable in size to \(K\).

Solution:

The carrying capacity is \(K=50000\), and the general solution of the logistic equation is

\[ N(t) = \frac{K}{1+Ae^{-rt}} = \frac{50000}{1+Ae^{-rt}}. \]

Use \(N(0)=5000\) to find \(A\):

\[ 5000 = \frac{50000}{1+A} \quad\Longrightarrow\quad 1+A = 10 \quad\Longrightarrow\quad A=9. \]

So \(N(t) = \dfrac{50000}{1+9e^{-rt}}\). Now use \(N(1)=8000\) to find \(r\):

\[ 8000 = \frac{50000}{1+9e^{-r}} \quad\Longrightarrow\quad 1+9e^{-r} = \frac{50000}{8000}=\frac{25}{4}. \]

Solving for \(e^{-r}\):

\[ 9e^{-r} = \frac{25}{4}-1 = \frac{21}{4} \quad\Longrightarrow\quad e^{-r} = \frac{21}{36}=\frac{7}{12}. \]

Since \(7/12<1\), this gives \(r = \ln\!\left(\dfrac{12}{7}\right) \approx 0.539 > 0\), a positive growth rate, as it should be. Thus

\[ \boxed{N(t) = \frac{50000}{1+9e^{-rt}}, \qquad r = \ln\!\left(\frac{12}{7}\right).} \]

Solution:

Equilibria occur where \(\frac{dP}{dt}=0\). Expanding the right-hand side,

\[ P(8-0.008P)-1500 = -0.008P^2+8P-1500 = 0. \]

Multiplying through by \(-125\) (chosen to clear the decimal, since \(1/0.008=125\)) turns this into a monic quadratic with integer coefficients:

\[ P^2 - 1000P + 187500 = 0. \]

Applying the quadratic formula,

\[ P = \frac{1000\pm\sqrt{1000^2-4(187500)}}{2} = \frac{1000\pm\sqrt{1{,}000{,}000-750{,}000}}{2} = \frac{1000\pm\sqrt{250{,}000}}{2}. \]

Since \(\sqrt{250{,}000}=500\),

\[ P = \frac{1000\pm500}{2}. \]

So the two equilibrium population levels are

\[ \boxed{P = 250 \quad \text{and} \quad P=750.} \]

(As a check, note that harvesting shrinks the "band" of equilibria inward from \(0\) and \(1000\) toward each other — consistent with the remark in the lecture notes that harvesting moves the critical points away from \(P=0\) and \(P=K\).)

Solution:

We examine the sign of \(\frac{dP}{dt}=-0.008(P-250)(P-750)\) on the three intervals the two equilibria create:

• For \(P<250\): both \((P-250)\) and \((P-750)\) are negative, so their product is positive; multiplying by \(-0.008\) makes \(\frac{dP}{dt}<0\) — the population is decreasing.
• For \(250 < P < 750\): \((P-250)>0\) but \((P-750)<0\), so the product is negative; multiplying by \(-0.008\) makes \(\frac{dP}{dt}>0\) — the population is increasing.
• For \(P>750\): both factors are positive, so the product is positive, and \(\frac{dP}{dt}<0\) — the population is decreasing.

At \(P=250\): just below, the population decreases further (moving away, toward \(0\) and extinction); just above, the population increases (also moving away, up toward \(750\)). Since solutions move away from \(250\) on both sides, it is an unstable equilibrium. Ecologically, any population that dips even slightly below \(250\) will be driven to extinction rather than recovering — it is a threshold, not a resting point.

At \(P=750\): just below, the population increases toward \(750\); just above, the population decreases back toward \(750\). Since solutions approach \(750\) from both sides, it is an asymptotically stable equilibrium — a population near \(750\) will settle there and can be sustainably harvested at this rate indefinitely.

Solution:

The general decay-series system from the lecture notes is

\[ \frac{dx}{dt} = -\lambda_1 x, \qquad \frac{dy}{dt} = -\lambda_2 y+\lambda_1 x, \qquad \frac{dz}{dt} = \lambda_2 y. \]

Substituting \(\lambda_1=0.5\) and \(\lambda_2=0.1\) gives the model

\[ \boxed{ \frac{dx}{dt} = -0.5x, \qquad \frac{dy}{dt} = -0.1y+0.5x, \qquad \frac{dz}{dt} = 0.1y, } \]

with initial conditions \(x(0)=100\), \(y(0)=0\), \(z(0)=0\).

Solution:

First check that each tank's volume is actually constant, since that is what will let us write "concentration \(=\) salt \(\div\) \(50\)" without worrying about a changing denominator. Tank A gains \(4\) (fresh) \(+1\) (from B) \(=5\) gal/min and loses \(5\) gal/min (to B) — net \(0\). Tank B gains \(5\) gal/min (from A) and loses \(1+4=5\) gal/min (back to A and out of the system) — net \(0\). So both volumes stay fixed at \(50\) gallons for all \(t\), and the concentrations are simply \(x(t)/50\) and \(y(t)/50\).

For Tank A: salt enters from the fresh water (concentration \(0\)) and from Tank B at \(1\) gal/min, and leaves toward Tank B at \(5\) gal/min: \[ \frac{dx}{dt} = 1\cdot\frac{y}{50} - 5\cdot\frac{x}{50} = \frac{y}{50}-\frac{x}{10}. \]

For Tank B: salt enters from Tank A at \(5\) gal/min, and leaves at a combined rate of \(1+4=5\) gal/min (back to A and out of the system): \[ \frac{dy}{dt} = 5\cdot\frac{x}{50} - 5\cdot\frac{y}{50} = \frac{x}{10}-\frac{y}{10}. \]

The mathematical model is

\[ \boxed{ \frac{dx}{dt} = -\frac{x}{10}+\frac{y}{50}, \qquad \frac{dy}{dt} = \frac{x}{10}-\frac{y}{10}, } \]

with initial conditions \(x(0)=20\), \(y(0)=15\).

Solution:

This problem is harder than the previous one because the two connecting flow rates (\(3\) gal/min and \(2\) gal/min) are different, so — unlike Problem 11 — the volumes in each tank change over time, and we need to track that before we can even write down the concentrations.

Tank A loses \(3\) gal/min (to B) and gains \(2\) gal/min (from B), a net loss of \(1\) gal/min, so starting from \(100\) gallons, \[ V_A(t) = 100 - t. \] Tank B gains \(3\) gal/min (from A) and loses \(2\) gal/min (to A), a net gain of \(1\) gal/min, so \[ V_B(t) = 100 + t. \]

Because each tank is well-stirred, the concentration in each is its salt content divided by its (now time-dependent) volume: \[ \text{concentration in A} = \frac{x(t)}{100-t}, \qquad \text{concentration in B} = \frac{y(t)}{100+t}. \]

For Tank A: salt enters from Tank B at \(2\) gal/min, and leaves toward Tank B at \(3\) gal/min: \[ \frac{dx}{dt} = 2\cdot\frac{y}{100+t} - 3\cdot\frac{x}{100-t}. \] For Tank B: salt enters from Tank A at \(3\) gal/min, and leaves toward Tank A at \(2\) gal/min: \[ \frac{dy}{dt} = 3\cdot\frac{x}{100-t} - 2\cdot\frac{y}{100+t}. \]

The mathematical model is

\[ \boxed{ \frac{dx}{dt} = \frac{2y}{100+t}-\frac{3x}{100-t}, \qquad \frac{dy}{dt} = \frac{3x}{100-t}-\frac{2y}{100+t}, } \]

with initial conditions \(x(0)=100\), \(y(0)=50\).

Solution:

The concentration term \(\dfrac{x(t)}{100-t}\) is only meaningful as long as \(100-t>0\), i.e. \(t<100\); at \(t=100\), Tank A's volume would reach zero, and beyond that the formula \(V_A(t)=100-t\) would predict a negative volume, which is physically impossible — the tank would have run dry and the well-stirred-mixture assumption (and the entire derivation of the differential equation) simply stops applying once that happens.

This is not a defect in the differential equation or its solution — the mathematics is perfectly correct on \(0\le t<100\); it is a reflection of the fact that the physical scenario we chose to model (draining faster than refilling) only makes sense for a finite window of time. The model was only ever an accurate description of the tank system while the tank actually contains liquid, and no mathematical technique can extend a model past the point where its own physical assumptions break down.

In Problem 11, the inflow and outflow rates at each tank were exactly balanced, so the volumes \(V_A(t)=V_B(t)=50\) stayed constant for all \(t\ge 0\) — there was never a volume that could hit zero, so that model remains valid for all time and needs no such domain restriction.

Solution:

Compute the Wronskian of \(y_1=e^{2x}\) and \(y_2=e^{-3x}\), using \(y_1'=2e^{2x}\) and \(y_2'=-3e^{-3x}\):

\[ W(y_1,y_2) = \begin{vmatrix} e^{2x} & e^{-3x} \\ 2e^{2x} & -3e^{-3x} \end{vmatrix} = e^{2x}(-3e^{-3x}) - e^{-3x}(2e^{2x}) = -3e^{-x}-2e^{-x} = -5e^{-x}. \]

Since \(-5e^{-x}\neq 0\) for every real \(x\), the Wronskian is never zero, so \(\{y_1,y_2\}\) is linearly independent, and since both functions solve the second-order homogeneous equation, this makes \(\{y_1,y_2\}\) a fundamental set of solutions on \((-\infty,\infty)\). Therefore the general solution is \[ \boxed{y = c_1e^{2x}+c_2e^{-3x}.} \]

Solution:

Recall the Pythagorean identity \(\sin^2(x)+\cos^2(x)=1\). Multiplying both sides by \(3\),

\[ 3 = 3\sin^2(x)+3\cos^2(x), \]

which we can rearrange into

\[ 1\cdot 3 - 3\cdot\sin^2(x) - 3\cdot\cos^2(x) = 0. \]

This is a linear combination of the three given functions, with coefficients \(1\), \(-3\), and \(-3\) — not all zero — that equals zero for every \(x\). By definition, this means the functions are \[ \boxed{\textbf{linearly dependent}.} \]

Solution:

Since we need to construct the equation rather than solve one, we work backward: compute the derivatives of \(y=\sin(2x)\) and look for a linear combination of \(y\) and its derivatives that vanishes identically.

\[ y = \sin(2x), \qquad y' = 2\cos(2x), \qquad y'' = -4\sin(2x). \]

Notice that \(y''=-4\sin(2x)=-4y\), so \(y''+4y=0\) is satisfied automatically for every \(x\): \[ y''+4y = -4\sin(2x)+4\sin(2x) = 0. \]

So a second-order, constant-coefficient, homogeneous differential equation having \(\sin(2x)\) as a solution is

\[ \boxed{y''+4y=0.} \]

(This makes sense from the Section 4.3 perspective too: the auxiliary equation \(m^2+4=0\) has roots \(m=\pm 2i\), giving general solution \(y=c_1\cos(2x)+c_2\sin(2x)\), which indeed contains \(\sin(2x)\) as the special case \(c_1=0,c_2=1\).)

Solution:

With three functions we need the \(3\times 3\) Wronskian, built from each function and its first two derivatives:

\[ f_1=1,\ f_1'=0,\ f_1''=0; \qquad f_2=e^x,\ f_2'=e^x,\ f_2''=e^x; \qquad f_3=e^{2x},\ f_3'=2e^{2x},\ f_3''=4e^{2x}. \] \[ W(f_1,f_2,f_3) = \begin{vmatrix} 1 & e^x & e^{2x} \\ 0 & e^x & 2e^{2x} \\ 0 & e^x & 4e^{2x} \end{vmatrix}. \]

Since the first column has only one nonzero entry (the \(1\) in the top row), we can expand along it, which reduces the determinant to a single \(2\times 2\) piece:

\[ W = 1\cdot\begin{vmatrix} e^x & 2e^{2x} \\ e^x & 4e^{2x} \end{vmatrix} = e^x(4e^{2x}) - 2e^{2x}(e^x) = 4e^{3x}-2e^{3x} = 2e^{3x}. \]

Since \(2e^{3x}\neq 0\) for every real \(x\), the Wronskian is never zero, so \[ \boxed{f_1=1,\ f_2=e^x,\ f_3=e^{2x}\ \text{are linearly independent on } (-\infty,\infty).} \]

Solution:

No. Two functions \(f\) and \(f'\) are linearly dependent precisely when \(f'\) is a constant multiple of \(f\), and plenty of functions satisfy this. The simplest example is \[ f(x) = e^x, \qquad f'(x) = e^x. \]

Here \(f'(x)-f(x)=0\) for every \(x\) — a nontrivial linear combination (coefficients \(1\) and \(-1\), not both zero) that vanishes identically — so by definition \(f\) and \(f'\) are linearly dependent in this case, not independent.

This is consistent with the Wronskian test: for \(f_1=f\), \(f_2=f'\), \[ W(f_1,f_2) = f_1f_2'-f_2f_1' = f\cdot f''-f'\cdot f' = e^x\cdot e^x - e^x\cdot e^x = 0 \] for \(f=e^x\), confirming dependence. (For a function like \(f(x)=x^2\), by contrast, \(f'=2x\) is not a constant multiple of \(f\), and one can check the corresponding Wronskian is nonzero away from \(x=0\), so in that case \(f\) and \(f'\) genuinely are linearly independent.) So the answer depends entirely on the specific function — there is no universal rule forcing independence.

Solution:

Step 1 (Identify \(P(x)\) and the known solution).

The equation is already in standard form \(y''+P(x)y'+Q(x)y=0\) with \(P(x)=4\) (a constant), and we are given \(y_1=e^{-2x}\).

Step 2 (Apply the reduction-of-order formula).

\[ y_2 = y_1\int \frac{e^{-\int P\,dx}}{y_1^2}\,dx = e^{-2x}\int \frac{e^{-4x}}{e^{-4x}}\,dx. \]

Step 3 (Simplify and integrate).

The integrand simplifies to \(1\), so

\[ y_2 = e^{-2x}\int 1\,dx = xe^{-2x}. \]

Step 4 (Write the general solution).

\[ \boxed{y = c_1e^{-2x}+c_2xe^{-2x}.} \]

(This matches what Section 4.3 would predict directly: the auxiliary equation \(m^2+4m+4=(m+2)^2=0\) has the repeated root \(m=-2\), giving exactly \(y=c_1e^{-2x}+c_2xe^{-2x}\).)

Solution:

Step 1 (Put the equation in standard form).

Divide by \(x^2\) to identify \(P(x)\):

\[ y'' + \frac{1}{x}y' = 0 \quad\Longrightarrow\quad P(x)=\frac{1}{x}. \]

Step 2 (Set up the reduction-of-order formula).

\[ y_2 = y_1\int \frac{e^{-\int P(x)\,dx}}{y_1^2}\,dx, \qquad \int P(x)\,dx = \int\frac{1}{x}\,dx = \ln x, \qquad e^{-\ln x}=\frac{1}{x}. \]

So

\[ y_2 = \ln(x)\int \frac{1}{x(\ln x)^2}\,dx. \]

Step 3 (Evaluate the integral by substitution).

Let \(u=\ln x\), so \(du=\frac{1}{x}dx\), turning the integral into

\[ \int \frac{1}{u^2}\,du = -\frac{1}{u} = -\frac{1}{\ln x}. \]

Step 4 (Assemble \(y_2\) and the general solution).

Substituting back,

\[ y_2 = \ln(x)\left(-\frac{1}{\ln x}\right) = -1. \]

Since a nonzero constant multiple of a solution is still a solution, we may simply take \(y_2=1\) (indeed, \(y=1\) satisfies \(x^2y''+xy'=0\) directly, since \(y'=y''=0\)). Thus

\[ \boxed{y = c_1\ln(x)+c_2.} \]

(This is consistent with treating the equation as a Cauchy–Euler equation: trying \(y=x^m\) gives \(m(m-1)+m=m^2=0\), a repeated root \(m=0\), whose two independent solutions are \(x^0=1\) and \(x^0\ln x=\ln x\) — matching exactly.)

Solution:

Step 1 (Put the equation in standard form).

Divide by \(x^2\):

\[ y'' + \frac{3}{x}y' + \frac{1}{x^2}y = 0 \quad\Longrightarrow\quad P(x)=\frac{3}{x}. \]

Step 2 (Substitute \(y_2=u(x)y_1\) directly, rather than only using the formula).

With \(y_1=x^{-1}\), let \(y_2=ux^{-1}\). Then

\[ y_2' = u'x^{-1}-ux^{-2}, \qquad y_2'' = u''x^{-1}-2u'x^{-2}+2ux^{-3}. \]

Substituting into \(x^2y_2''+3xy_2'+y_2\) and simplifying term by term:

\[ x^2y_2'' = u''x - 2u'+2ux^{-1}, \qquad 3xy_2' = 3u'-3ux^{-1}, \qquad y_2 = ux^{-1}. \]

Adding these three pieces, the \(u\) terms cancel completely (as they must, since \(y_1\) itself solves the equation), leaving

\[ xu''+u' = 0. \]

Step 3 (Solve the reduced equation for \(w=u'\)).

With \(w=u'\): \(xw'+w=0\), i.e. \(\frac{w'}{w}=-\frac{1}{x}\), so \(\ln|w|=-\ln|x|+C\), giving

\[ w = \frac{1}{x}. \]

Step 4 (Integrate to find \(u\), then assemble \(y_2\)).

\[ u = \int \frac{1}{x}\,dx = \ln(x), \]

so \(y_2 = u\,y_1 = \dfrac{\ln x}{x}\). As a check, the Wronskian is

\[ W(y_1,y_2) = \frac{1}{x}\cdot\frac{1-\ln x}{x^2} - \frac{\ln x}{x}\cdot\left(-\frac{1}{x^2}\right) = \frac{1-\ln x+\ln x}{x^3} = \frac{1}{x^3} \ne 0, \]

confirming \(\{y_1,y_2\}\) is genuinely a fundamental set (and this matches Abel's formula \(W=e^{-\int P\,dx}=e^{-3\ln x}=x^{-3}\) exactly). So the general solution is

\[ \boxed{y = \frac{1}{x}\left(c_1+c_2\ln x\right).} \]

(This agrees with treating the equation as a Cauchy–Euler equation: \(m(m-1)+3m+1=(m+1)^2=0\) has the repeated root \(m=-1\), giving exactly \(y=x^{-1}(c_1+c_2\ln x)\).)

Solution:

Recall that a pair of solutions \(\{y_1,y_2\}\) of a second-order homogeneous linear equation forms a fundamental set exactly when their Wronskian is nonzero everywhere on the interval of interest.

The formula \(W(y_1,y_2)=e^{-\int P(x)\,dx}\) expresses the Wronskian directly as an exponential function. The exponential function \(e^{u}\) is never zero for any real (or even complex) value of \(u\) — no matter what \(P(x)\) is, as long as it is continuous so that \(\int P(x)\,dx\) is a well-defined real number at each point of the interval, \(e^{-\int P(x)\,dx}\) can never equal zero.

So the moment we know \(y_2\) was built by the reduction-of-order construction (rather than picked arbitrarily), we automatically know its Wronskian with \(y_1\) has this exponential form, and therefore is guaranteed to be nonzero — linear independence, and hence "fundamental set" status, comes for free from the construction itself, without ever needing to write out \(y_2\) explicitly or plug it into a determinant to check.

Solution:

The auxiliary equation is

\[ m^2-36=0 \quad\Longrightarrow\quad m=\pm 6, \]

two distinct real roots. So the general solution is

\[ \boxed{y=c_1e^{6x}+c_2e^{-6x}.} \]

Solution:

The auxiliary equation is

\[ m^2+9=0 \quad\Longrightarrow\quad m=\pm 3i, \]

a complex conjugate pair with \(\alpha=0\) and \(\beta=3\). So the general solution is

\[ \boxed{y=c_1\cos(3x)+c_2\sin(3x).} \]

Solution:

The auxiliary equation is

\[ m^2+12m+36 = (m+6)^2 = 0, \]

a repeated root \(m=-6\). So the general solution is

\[ y = c_1e^{-6x}+c_2xe^{-6x}. \]

Apply \(y(0)=2\):

\[ y(0) = c_1 = 2. \]

Differentiate to apply the second condition. By the product rule,

\[ y' = -6c_1e^{-6x} + c_2\left(e^{-6x}-6xe^{-6x}\right). \]

At \(x=0\):

\[ y'(0) = -6c_1+c_2 = 1 \quad\Longrightarrow\quad -12+c_2=1 \quad\Longrightarrow\quad c_2=13. \]

So the solution to the initial value problem is

\[ \boxed{y = 2e^{-6x}+13xe^{-6x}.} \]

Solution:

This is harder than the second-order cases because the auxiliary equation is now a cubic,

\[ m^3-2m^2-5m+6=0, \]

which does not factor by inspection the way a quadratic does — we first need to locate one root, typically by testing small integer candidates. Trying \(m=1\):

\[ 1-2-5+6 = 0. \checkmark \]

So \(m=1\) is a root, meaning \((m-1)\) is a factor. Dividing \(m^3-2m^2-5m+6\) by \((m-1)\) (synthetic or long division) gives

\[ m^3-2m^2-5m+6 = (m-1)(m^2-m-6) = (m-1)(m-3)(m+2). \]

So the three roots are \(m=1,\ 3,\ -2\) — all real and distinct. By the extension of the distinct-real-roots case to higher order, the general solution is

\[ \boxed{y = c_1e^{x}+c_2e^{3x}+c_3e^{-2x}.} \]

Solution:

The quadratic formula \(m=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) shows that everything hinges on the quantity under the square root:

• If \(b^2-4ac>0\), the square root is a positive real number, so the two roots \(m_1\neq m_2\) are real and distinct, giving two genuinely different exponentials \(e^{m_1x}\) and \(e^{m_2x}\) that combine additively — no oscillation, just growth or decay.
• If \(b^2-4ac=0\), the square root vanishes, collapsing both roots onto the single value \(m=-b/(2a)\); reduction of order is then needed to find the second solution, which turns out to be \(xe^{mx}\) rather than a second independent exponential.
• If \(b^2-4ac<0\), the square root is the square root of a negative number, forcing the two roots to be a complex-conjugate pair \(\alpha\pm i\beta\); Euler's formula converts these into the oscillatory pair \(e^{\alpha x}\cos(\beta x)\) and \(e^{\alpha x}\sin(\beta x)\).

For \(y''+2y'+5y=0\): \(a=1\), \(b=2\), \(c=5\), so

\[ b^2-4ac = 4-20 = -16 < 0, \]

which puts us in the third case: the solution is oscillatory. Without solving further, we can also read off the "envelope" behavior directly from \(\alpha=-b/(2a)=-1\): since the oscillation is multiplied by \(e^{\alpha x}=e^{-x}\), and \(\alpha<0\), the oscillation's amplitude decays toward zero as \(x\to\infty\) (a damped oscillation), rather than growing or staying constant.

Solution:

Step 1 (Solve the associated homogeneous equation).

\[ m^2+m-6 = (m+3)(m-2) = 0 \quad\Longrightarrow\quad m=-3,\ 2, \qquad y_c = c_1e^{-3x}+c_2e^{2x}. \]

Step 2 (Guess the form of \(y_p\)).

Since \(g(x)=2x\) is a first-degree polynomial (and neither \(1\) nor \(x\) shows up in \(y_c\), so no duplication to worry about), guess

\[ y_p = Ax+B, \qquad y_p'=A, \qquad y_p''=0. \]

Step 3 (Substitute and match coefficients).

\[ y_p''+y_p'-6y_p = 0+A-6(Ax+B) = -6Ax+(A-6B) \stackrel{\text{set}}{=} 2x+0. \]

Matching the \(x\)-coefficients and the constants separately:

\[ -6A=2 \;\Longrightarrow\; A=-\frac13, \qquad A-6B=0 \;\Longrightarrow\; -\frac13-6B=0 \;\Longrightarrow\; B=-\frac{1}{18}. \]

Step 4 (Assemble the general solution).

\[ \boxed{y = c_1e^{-3x}+c_2e^{2x} - \frac13x-\frac{1}{18}.} \]

Solution:

Step 1 (Solve the associated homogeneous equation).

\[ m^2-16=0 \quad\Longrightarrow\quad m=\pm 4, \qquad y_c = c_1e^{4x}+c_2e^{-4x}. \]

Step 2 (Guess the form of \(y_p\)).

Since \(g(x)=2e^{3x}\) and \(e^{3x}\) does not already appear in \(y_c\) (the homogeneous exponentials are \(e^{4x}\) and \(e^{-4x}\), not \(e^{3x}\)), guess

\[ y_p = Ae^{3x}, \qquad y_p'=3Ae^{3x}, \qquad y_p''=9Ae^{3x}. \]

Step 3 (Substitute and match coefficients).

\[ y_p''-16y_p = 9Ae^{3x}-16Ae^{3x} = -7Ae^{3x} \stackrel{\text{set}}{=} 2e^{3x} \quad\Longrightarrow\quad A=-\frac27. \]

Step 4 (Assemble the general solution).

\[ \boxed{y = c_1e^{4x}+c_2e^{-4x}-\frac{2}{7}e^{3x}.} \]

Solution:

Step 1 (Solve the associated homogeneous equation).

\[ m^2-4m+4=(m-2)^2=0 \quad\Longrightarrow\quad m=2 \text{ (repeated)}, \qquad y_c = c_1e^{2x}+c_2xe^{2x}. \]

Step 2 (Notice — and fix — a duplication).

The naive guess for \(g(x)=6e^{2x}\) would be \(y_p=Ae^{2x}\), but \(e^{2x}\) already appears in \(y_c\) — and worse, so does \(xe^{2x}\), since \(m=2\) is a repeated root. Per the remark in the lecture notes, we must multiply our guess by \(x\) enough times to clear every duplication, so here we need \(x^2\) rather than just \(x\): \[ y_p = Ax^2e^{2x}. \]

Step 3 (Differentiate and substitute).

By the product rule,

\[ y_p' = Ae^{2x}(2x+2x^2), \qquad y_p'' = Ae^{2x}(2+8x+4x^2). \]

Substituting into the left-hand side,

\[ y_p''-4y_p'+4y_p = Ae^{2x}\Big[(2+8x+4x^2) - 4(2x+2x^2) + 4x^2\Big]. \]

Collecting like powers of \(x\) inside the brackets: the \(x^2\) terms are \(4-8+4=0\), the \(x\) terms are \(8-8=0\), leaving only the constant \(2\):

\[ y_p''-4y_p'+4y_p = 2Ae^{2x}. \]

Step 4 (Match coefficients and assemble).

\[ 2Ae^{2x} \stackrel{\text{set}}{=} 6e^{2x} \quad\Longrightarrow\quad A=3. \]

So \(y_p=3x^2e^{2x}\), and the general solution is

\[ \boxed{y = c_1e^{2x}+c_2xe^{2x}+3x^2e^{2x}.} \]

Solution:

Undetermined coefficients does not apply here, since \(\sec(x)\) is not a polynomial, exponential, sine, or cosine (its own derivatives never cycle back through a finite family of functions the way \(\sin\), \(\cos\), or \(e^{mx}\) do), so we need the more general method of variation of parameters.

Step 1 (Find \(y_c\) and identify \(y_1,y_2,f(x)\)).

\[ m^2+1=0 \quad\Longrightarrow\quad m=\pm i, \qquad y_c = c_1\cos x+c_2\sin x. \]

So \(y_1=\cos x\), \(y_2=\sin x\), and \(f(x)=\sec x\).

Step 2 (Compute the Wronskians).

\[ W(y_1,y_2) = \begin{vmatrix}\cos x & \sin x \\ -\sin x & \cos x\end{vmatrix} = \cos^2x+\sin^2x = 1. \] \[ W_1 = \begin{vmatrix}0 & \sin x \\ \sec x & \cos x\end{vmatrix} = -\sin x\sec x = -\tan x, \qquad W_2 = \begin{vmatrix}\cos x & 0 \\ -\sin x & \sec x\end{vmatrix} = \cos x\sec x = 1. \]

Step 3 (Solve for \(u_1',u_2'\) and integrate).

\[ u_1' = \frac{W_1}{W} = -\tan x, \qquad u_2' = \frac{W_2}{W} = 1. \]

Integrating (using \(\int\tan x\,dx=-\ln|\cos x|\), so \(\int-\tan x\,dx=\ln|\cos x|\)):

\[ u_1 = \int -\tan x\,dx = \ln|\cos x|, \qquad u_2 = \int 1\,dx = x. \]

Step 4 (Assemble \(y_p\) and the general solution).

\[ y_p = u_1y_1+u_2y_2 = \cos(x)\ln|\cos x| + x\sin x. \]

So

\[ \boxed{y = c_1\cos x+c_2\sin x+\cos(x)\ln|\cos x|+x\sin x.} \]

Solution:

Undetermined coefficients works by guessing a form for \(y_p\) built out of the same "family" of functions as \(g(x)\) — polynomials, exponentials, and sines/cosines all have the special property that differentiating them (possibly several times) only ever produces more functions from that same finite family, so a guess with a handful of unknown coefficients can always be made to match.

The function \(\sec(x)\) does not have this property: its derivatives introduce new, different combinations of \(\sec x\) and \(\tan x\) at every step (\(\sec x\), then \(\sec x\tan x\), then \(\sec x\tan^2x+\sec^3x\), and so on) rather than cycling back through a small closed family the way \(\sin x\) and \(\cos x\) do. There is no finite list of terms we could write down for \(y_p\) with undetermined coefficients that would be guaranteed to reproduce \(\sec(x)\) after substitution — the method simply has no entry point here.

Variation of parameters sidesteps this issue entirely, because it never tries to guess the form of \(y_p\) at all. Instead it only needs a fundamental set \(y_1,y_2\) of the homogeneous equation (which we always have from Section 4.3 for constant coefficients) and then directly integrates expressions built from \(y_1\), \(y_2\), and \(f(x)\). This procedure works for any continuous \(f(x)\) — the only thing that can go wrong is that the resulting integrals \(\int u_1'\,dx\) and \(\int u_2'\,dx\) might not have a closed-form antiderivative, but for \(f(x)=\sec(x)\) they do (as the previous problem shows), so the method goes through without difficulty.