Practice / Review Problems — MATH 2410

Solution:

To verify a proposed solution, we substitute it into the differential equation and check that the equation becomes an identity. First compute \(y'\):

\[ y' = \frac{d}{dx} e^{-x/2} = -\frac{1}{2} e^{-x/2}. \]

Now substitute both \(y\) and \(y'\) into the left-hand side of the differential equation:

\[ 2y' + y = 2\left(-\frac{1}{2} e^{-x/2}\right) + e^{-x/2} = -e^{-x/2}+e^{-x/2} = 0. \]

Since this matches the right-hand side of the equation for every \(x\), and \(y=e^{-x/2}\) is continuous and differentiable on all of \((-\infty,\infty)\), we conclude that \(y = e^{-x/2}\) is indeed a solution on \((-\infty, \infty)\).

Solution:

• Type: Ordinary differential equation, since only ordinary derivatives of \(y\) (with respect to the single variable \(x\)) appear.
• Order: Second order, because the highest derivative present is \(y''\).
• Linearity: Linear. Even though the coefficient \(5\tan(x)\) in front of \(y'\) is a complicated function of \(x\), what matters for linearity is how \(y\) and its derivatives appear — here \(y\), \(y'\), and \(y''\) each appear only to the first power, are not multiplied by each other, and are not plugged into any other function (like \(\sin(y)\) or \(y^2\)).

Solution:

• Type: Ordinary differential equation.
• Order: First order, since the highest derivative is \(y'\).
• Linearity: Nonlinear, because the term \(y^2\) involves the unknown function raised to a power greater than one. A single such term is enough to make the whole equation nonlinear, regardless of how simple the rest of the equation is.

Solution:

• Type: Partial differential equation, since \(u\) depends on two variables (\(x\) and \(t\)) and partial derivatives with respect to both appear.
• Order: Second order, because the highest derivative is \(\frac{\partial^2 u}{\partial x^2}\).
• Linearity: Linear, since \(u\) and its partial derivatives appear only to the first power and are not multiplied together.

Solution:

This family is given implicitly — \(y\) is not solved for in terms of \(x\) — so we differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\) and using the chain rule on the \(y^2\) term:

\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(c) \quad\Longrightarrow\quad 2x + 2y \frac{dy}{dx} = 0. \]

Solving this for \(y'\):

\[ 2y\frac{dy}{dx} = -2x \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{x}{y}. \]

This is exactly the given differential equation, so the family satisfies it. Note the family only defines \(y\) implicitly, and the derivative \(y'=-x/y\) is undefined when \(y=0\); the solution circles \(x^2+y^2=c\) must therefore be split into upper and lower semicircular arcs (where \(y\neq 0\)) to be viewed as genuine functions \(y(x)\).

Solution:

Differentiate \(y\) with respect to \(x\):

\[ y' = ce^{x} - 1. \]

Substitute \(y\) and \(y'\) into the left-hand side of the differential equation:

\[ y' - y = (ce^x - 1) - (ce^x - x - 1) = ce^x - 1 - ce^x + x + 1 = x. \]

The \(ce^x\) terms cancel completely, regardless of the value of \(c\) — this is exactly what it means for \(y=ce^x-x-1\) to be a whole family of solutions rather than a single one: every choice of \(c\) produces a different curve, but all of them satisfy \(y'-y=x\).

Solution:

This example is harder because \(y\) is written as a product of \(\sqrt{x}\) with an integral whose upper limit is the variable \(x\). To differentiate it we need both the product rule and the Fundamental Theorem of Calculus (which tells us \(\frac{d}{dx}\int_4^x g(t)\,dt = g(x)\)). Applying the product rule,

\[ y' = \frac{d}{dx}\left(\sqrt{x}\right) \int_4^x \frac{\cos(t)}{\sqrt{t}}\,dt \;+\; \sqrt{x}\cdot \frac{d}{dx}\int_4^x \frac{\cos(t)}{\sqrt{t}}\,dt. \]

The first factor's derivative is \(\frac{1}{2\sqrt{x}}\), and by the Fundamental Theorem of Calculus the second derivative is just the integrand evaluated at \(t=x\), namely \(\frac{\cos(x)}{\sqrt{x}}\). So

\[ y' = \frac{1}{2\sqrt{x}} \int_4^x \frac{\cos(t)}{\sqrt{t}}\, dt + \sqrt{x}\cdot\frac{\cos(x)}{\sqrt{x}} = \frac{1}{2\sqrt{x}} \int_4^x \frac{\cos(t)}{\sqrt{t}}\, dt + \cos(x). \]

Now multiply both sides by \(2x\), which is the coefficient of \(y'\) in the differential equation:

\[ 2xy' = 2x\cos(x) + \sqrt{x}\int_4^x \frac{\cos(t)}{\sqrt{t}}\, dt = 2x\cos(x) + y, \]

where in the last step we recognized \(\sqrt{x}\int_4^x \frac{\cos(t)}{\sqrt{t}}\,dt\) as exactly \(y\) itself. Rearranging,

\[ 2xy' - y = 2x\cos(x), \]

which is the original differential equation, so \(y\) is indeed a solution.

Solution:

Linearity is a statement about how the unknown function \(y\) and its derivatives appear in the equation — not about how complicated the coefficients in front of them are. A differential equation is linear as long as it can be written in the form \[ a_n(x) y^{(n)} + \cdots + a_1(x) y' + a_0(x) y = g(x), \] where the coefficients \(a_i(x)\) are allowed to be any functions of \(x\) alone — they can be as complicated as we like (polynomials, trig functions, exponentials, etc.), as long as they never multiply \(y\) by \(y'\), never raise \(y\) or its derivatives to a power other than one, and never plug \(y\) into another function like \(\sin(y)\) or \(e^y\).

In \(x^2y' + \cos(x)y = e^x\), the coefficients \(a_1(x)=x^2\) and \(a_0(x)=\cos(x)\) are both functions of \(x\) only. The unknown function \(y\) and its derivative \(y'\) each appear to the first power, with no products of \(y\) and \(y'\) and no \(y\) inside another function. So despite looking intimidating, this equation is linear.

Solution:

Differentiate:

\[ y' = -2ce^{-2x}. \]

Substitute into the differential equation:

\[ y' + 2y = -2ce^{-2x} + 2ce^{-2x} = 0, \]

so \(y=ce^{-2x}\) is a solution for every \(c\). Applying the initial condition \(y(0)=5\):

\[ y(0) = ce^{0} = c = 5, \]

so the particular solution is

\[ \boxed{y = 5e^{-2x}.} \]

Solution:

We begin by computing the derivative of \[ y = \frac{1}{x^2 + c} = (x^2 + c)^{-1}. \]

Using the chain rule, \[ y' = -1(x^2 + c)^{-2}(2x) = -\frac{2x}{(x^2 + c)^2}. \]

Next, compute the term \(2xy^2\): \[ y^2 = \left(\frac{1}{x^2 + c}\right)^2 = \frac{1}{(x^2 + c)^2}, \] so \[ 2xy^2 = \frac{2x}{(x^2 + c)^2}. \]

Substitute into the differential equation: \[ y' + 2xy^2 = -\frac{2x}{(x^2 + c)^2} + \frac{2x}{(x^2 + c)^2} = 0. \]

Therefore, \(y = \frac{1}{x^2 + c}\) is a solution for every constant \(c\).

Now apply the initial condition \(y(2) = \frac{1}{3}\): \[ \frac{1}{2^2 + c} = \frac{1}{3} \Rightarrow 4 + c = 3 \Rightarrow c = -1. \]

The particular solution is \[ y = \frac{1}{x^2 - 1}. \]

This solution is undefined when \(x^2 - 1 = 0\), i.e., at \(x = \pm 1\). The domain \((-\infty,\infty)\) is therefore split into three pieces by these two points: \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\). Since the initial condition occurs at \(x = 2\), we need the piece that contains \(2\), which is \[ (1, \infty). \]

Solution:

We write the differential equation in the form \[ y' = f(x,y), \quad \text{where} \quad f(x,y) = \sqrt{xy}. \]

For the existence and uniqueness theorem to apply, we need to find a rectangle around \((x_0,y_0)\) on which both \(f(x,y)\) and \(\frac{\partial f}{\partial y}\) are continuous.

First, observe that \(\sqrt{xy}\) is only real-valued (hence only continuous, as opposed to undefined) when \[ xy \ge 0. \]

Next, compute the partial derivative with respect to \(y\): \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (xy)^{1/2} = \frac{1}{2}(xy)^{-1/2} \cdot x = \frac{x}{2\sqrt{xy}}. \]

This expression has \(\sqrt{xy}\) in the denominator, so it is undefined when \(xy = 0\) (division by zero) and not real when \(xy < 0\) (same issue as before). So \(\frac{\partial f}{\partial y}\) is continuous only on the strictly stronger condition \[ xy > 0. \]

One region satisfying \(xy>0\) — and the natural one to pick, since it is an open rectangle-friendly region — is \[ x > 0 \quad \text{and} \quad y > 0. \] (The region \(x<0,\ y<0\) also satisfies \(xy>0\) and would work equally well.)

In this region, both hypotheses of the existence and uniqueness theorem are satisfied, so there exists a unique solution whose graph passes through any point \((x_0, y_0)\) with \(x_0 > 0\) and \(y_0 > 0\).

Solution:

First, \(y=0\) is a constant function, so \(y'=0\), and the right-hand side is \(3(0)^{2/3}=0\) — the equation holds, and \(y(0)=0\) is satisfied.

For \(y=x^3\): \(y'=3x^2\), and the right-hand side is \[ 3y^{2/3} = 3(x^3)^{2/3} = 3x^2 \] (squaring first removes the sign ambiguity, so this holds for every real \(x\), positive or negative). So \(y'=3x^2\) matches the right-hand side, and \(y(0)=0^3=0\) is satisfied as well. Both functions genuinely solve the IVP.

This does not contradict the theorem because the theorem is a sufficient condition for uniqueness, not a necessary one — it guarantees uniqueness only where its hypotheses actually hold. Here, \(f(x,y)=3y^{2/3}\), so \[ \frac{\partial f}{\partial y} = 2y^{-1/3}, \] which blows up (is discontinuous) exactly at \(y=0\) — precisely the point our initial condition sits on. Since the hypothesis "\(\partial f/\partial y\) continuous near \((x_0,y_0)\)" fails at \((0,0)\), the theorem simply does not apply there, and it should be no surprise that uniqueness actually fails at that point.

Solution:

Let \(x(t)\) denote the amount of drug in the bloodstream at time \(t\), so once again \[ \frac{dx}{dt} = (\text{rate in}) - (\text{rate out}). \]

The drug is infused at a constant rate \(r\), so \[ \text{Rate in} = r. \]

The drug is removed at a rate proportional to the amount present, so \[ \text{Rate out} = kx, \] where \(k > 0\) is a proportionality constant.

Substituting into the net-rate equation gives \[ \boxed{\frac{dx}{dt} = r - kx.} \] This is a first-order linear differential equation commonly used in pharmacokinetics.

Solution:

"Rate proportional to the amount present" translates directly to \(k\cdot A\) for some constant \(k\). Since the substance is decaying, the amount is decreasing, which corresponds to \(k<0\). The initial condition just restates "initially \(A_0\) grams" as \(A(0)=A_0\). Altogether, \[ \boxed{\frac{dA}{dt} = kA, \qquad A(0) = A_0, \qquad k<0.} \]

Solution:

Let \(P(t)\) denote the population at time \(t\). The overall strategy for any "net rate" modeling problem is the same: identify every rate flowing in, identify every rate flowing out, and write \[ \frac{d(\text{quantity})}{dt} = (\text{rate in}) - (\text{rate out}). \]

Here, births add to the population and deaths remove from it, so births play the role of "rate in" and deaths play the role of "rate out." The birth rate is proportional to the population, so we model it as \[ \text{Birth rate} = aP, \] where \(a > 0\) is a proportionality constant. The death rate is proportional to the square of the population, so it is modeled as \[ \text{Death rate} = bP^2, \] where \(b > 0\) is a proportionality constant.

The net rate of change of the population is then \[ \frac{dP}{dt} = (\text{birth rate}) - (\text{death rate}) = aP - bP^2. \]

This is a first-order, nonlinear differential equation (because of the \(P^2\) term) known as the logistic equation — it is the standard model for a population growing in an environment with limited resources.

Solution:

Let \(A(t)\) be the amount of salt (in pounds) in the tank at time \(t\). As with any mixing-tank problem, the rate of change of the amount of salt is \[ \frac{dA}{dt} = (\text{rate in}) - (\text{rate out}). \]

Rate in: The incoming solution has concentration \(2\) lb/gal and enters at \(3\) gal/min, so the salt enters at \[ \text{Rate in} = (2\ \text{lb/gal})(3\ \text{gal/min}) = 6\ \text{lb/min}. \]

Rate out is trickier here, because the inflow (\(3\) gal/min) and outflow (\(2\) gal/min) are different, so the volume of liquid in the tank is not constant — it grows at a net rate of \(3-2=1\) gal/min. Starting from \(300\) gallons, \[ V(t) = 300 + (3 - 2)t = 300 + t \ \text{gal}. \]

Because the mixture is well-stirred, the concentration of salt anywhere in the tank at time \(t\) is the same, namely the total salt divided by the total volume: \[ \text{concentration at time } t = \frac{A(t)}{300 + t} \ \text{lb/gal}. \]

Since the mixture leaves at \(2\) gal/min, the rate at which salt leaves the tank is that concentration times the outflow rate: \[ \text{Rate out} = 2 \cdot \frac{A(t)}{300 + t}. \]

Combining rate in and rate out gives the differential equation \[ \boxed{\frac{dA}{dt} = 6 - \frac{2A}{300 + t}.} \]

Solution:

This is the same modeling setup as the draining-tank example in Section 1.3, just with different numbers — so the strategy is identical: relate \(\frac{dV}{dt}\) to the outflow rate, relate \(V\) to \(h\) through the shape of the tank, then combine the two.

First, convert the hole's radius to feet: \(2\) inches \(=\frac{1}{6}\) ft, so the area of the hole is \[ A_h = \pi\left(\frac{1}{6}\right)^2 = \frac{\pi}{36}\ \text{ft}^2. \]

The volume is leaving the tank, so \(\frac{dV}{dt}\) is the negative of the outflow magnitude: \[ \frac{dV}{dt} = -cA_h\sqrt{2gh} = -(0.6)\left(\frac{\pi}{36}\right)\sqrt{(2)(32)h} = -(0.6)\left(\frac{\pi}{36}\right)(8\sqrt{h}). \]

Simplifying the constant: \[ (0.6)\left(\frac{\pi}{36}\right)(8) = \frac{4.8\pi}{36} = \frac{2\pi}{15}, \] so \[ \frac{dV}{dt} = -\frac{2\pi}{15}\sqrt{h}. \]

Since the tank is cylindrical with radius \(4\) ft, its cross-sectional area is constant: \(A_w = \pi(4)^2 = 16\pi\ \text{ft}^2\), and \(V(t)=A_wh(t)\), so \(\frac{dV}{dt} = A_w\frac{dh}{dt} = 16\pi\frac{dh}{dt}\). Combining this with the expression above, \[ 16\pi\frac{dh}{dt} = -\frac{2\pi}{15}\sqrt{h} \quad\Longrightarrow\quad \frac{dh}{dt} = -\frac{2\pi}{15\cdot 16\pi}\sqrt{h} = -\frac{1}{120}\sqrt{h}. \]

So the differential equation modeling the height of water is \[ \boxed{\frac{dh}{dt} = -\frac{\sqrt{h}}{120}.} \]

Solution:

Model (a) is the pure exponential growth model: the growth rate is always proportional to the current population, no matter how large \(P\) gets, so it implicitly assumes unlimited resources — space, food, etc. — forever.

Model (b), the logistic model, includes the extra factor \(\left(1-\frac{P}{N}\right)\), which is close to \(1\) (so it behaves like model (a)) when \(P\) is small compared to \(N\), but shrinks toward \(0\) as \(P\) approaches \(N\), and would even become negative if \(P\) exceeded \(N\). This builds in a carrying capacity \(N\): the maximum population the environment can sustain given limited resources. As the population approaches that capacity, the growth rate automatically slows down and levels off, rather than growing without bound the way model (a) does.

Solution:

Set \(y'=0\) and solve for the critical point:

\[ 4-2y=0 \quad\Longrightarrow\quad y=2. \]

We can also write \(y' = -2(y-2)\), which makes the sign of \(y'\) easy to read off:

• For \(y<2\): \((y-2)<0\), so \(y'=-2(y-2)>0\) — solutions increase toward \(y=2\).
• For \(y>2\): \((y-2)>0\), so \(y'=-2(y-2)<0\) — solutions decrease toward \(y=2\).

Since solutions approach \(y=2\) from both sides, the critical point \(y=2\) is asymptotically stable (an attractor).

Solution:

First, factor the right-hand side to see the critical points clearly:

\[ y' = y^2(1 - y). \]

Critical points occur when \( y' = 0 \):

\[ y = 0, \quad y = 1. \]

To classify each one, examine the sign of \(y'\) on the intervals it creates:

• For \(y < 0\): \(y^2 > 0\) and \(1-y > 0\), so \(y' > 0\) (increasing).
• For \(0 < y < 1\): \(y^2 > 0\) and \(1-y > 0\), so \(y' > 0\) (increasing).
• For \(y > 1\): \(y^2>0\) but \(1-y < 0\), so \(y' < 0\) (decreasing).

At \(y=0\): just below \(0\), solutions increase toward \(0\); just above \(0\), solutions increase away from \(0\). Approaching from one side but not the other makes \(y=0\) semi-stable.

At \(y=1\): just below \(1\), solutions increase toward \(1\); just above \(1\), solutions decrease toward \(1\). Approaching from both sides makes \(y=1\) asymptotically stable.

Solution:

Factor the quadratic factor:

\[ y' = y(y^2 - 6y + 8) = y(y-2)(y-4). \]

Critical points occur when \(y' = 0\):

\[ y = 0, \quad y = 2, \quad y = 4. \]

Sign analysis (checking one test point in each of the four intervals):

• \(y < 0\): all three factors \(y\), \((y-2)\), \((y-4)\) are negative, so their product is negative: \(y' < 0\).
• \(0 < y < 2\): \(y>0\), \((y-2)<0\), \((y-4)<0\): two negatives and one positive give \(y'>0\).
• \(2 < y < 4\): \(y>0\), \((y-2)>0\), \((y-4)<0\): \(y'<0\).
• \(y > 4\): all three factors positive, so \(y'>0\).

Reading off the behavior on each side of each critical point:

• \(y=0\): decreasing below, increasing above — moves away on both sides — unstable.
• \(y=2\): increasing below, decreasing above — approaches from both sides — asymptotically stable.
• \(y=4\): decreasing below, increasing above — moves away on both sides — unstable.

Solution:

Factor \(y\) out of the numerator:

\[ y' = \frac{y(e^y - 9)}{e^y}. \]

Since \(e^y > 0\) for every real \(y\), it never contributes a sign change (in the numerator or as the denominator), so the sign of \(y'\) is determined entirely by \(y(e^y-9)\).

Critical points occur when the numerator is zero:

\[ y = 0, \qquad e^y = 9 \Rightarrow y = \ln 9. \]

(Note \(\ln 9 \approx 2.197\), so \(0 < \ln 9\).)

Sign analysis:

• \(y < 0\): \(y<0\), and \(e^y<1<9\) so \(e^y-9<0\). Product of two negatives: \(y'>0\).
• \(0 < y < \ln 9\): \(y>0\), but \(e^y<9\) still (since \(y<\ln 9\)), so \(e^y-9<0\). Product of a positive and a negative: \(y'<0\).
• \(y > \ln 9\): \(y>0\) and now \(e^y>9\), so \(e^y-9>0\). Product of two positives: \(y'>0\).

Classification:

• \(y=0\): increasing below, decreasing above — approaches from both sides — asymptotically stable.
• \(y=\ln 9\): decreasing below, increasing above — moves away on both sides — unstable.

Solution:

A critical point \(y=c\) satisfies \(f(c)=0\), so the constant function \(y(x)=c\) is itself a solution to the differential equation.

If a nonconstant solution \(y(x)\) were to cross \(y=c\), then at the crossing point \(x_0\), both the constant solution and the nonconstant solution would pass through the same point \((x_0,c)\).

But the existence-and-uniqueness theorem guarantees that a first-order differential equation (with \(f\) and \(\partial f/\partial y\) continuous, which is typically the case for the nice functions \(f(y)\) we work with) has only one solution through any given point. Two different solutions cannot both pass through \((x_0,c)\).

Therefore, once a solution's graph reaches \(y=c\), it must in fact be the constant solution from that point on — crossing is impossible, which is exactly why solution curves of autonomous equations never cross equilibrium (constant) solutions.

Solution:

Step 1 (Separate variables).

\[ y\,dy = x\,dx. \]

Step 2 (Integrate both sides).

\[ \int y\,dy = \int x\,dx \quad\Longrightarrow\quad \frac{y^2}{2} = \frac{x^2}{2} + C. \]

Step 3 (Simplify).

Multiplying through by \(2\) and renaming the constant, the implicit general solution is

\[ \boxed{y^2 = x^2 + C.} \]

Solution:

Separate variables by dividing both sides by \(\cos^3(3x)\):

\[ 2y\, dy = \frac{-\sin(3x)}{\cos^3(3x)}\, dx. \]

It helps to rewrite the right side as a product of tangent and secant, since \(\frac{\sin(3x)}{\cos(3x)}=\tan(3x)\) and \(\frac{1}{\cos^2(3x)}=\sec^2(3x)\):

\[ 2y\, dy = -\tan(3x)\sec^2(3x)\, dx. \]

Integrate both sides:

\[ \int 2y\, dy = -\int \tan(3x)\sec^2(3x)\, dx. \]

The left side is immediate: \(\int 2y\,dy = y^2\). For the right side, use the substitution \(u=\tan(3x)\), so \(du = 3\sec^2(3x)\,dx\), i.e. \(\sec^2(3x)\,dx = \frac{1}{3}du\):

\[ \int \tan(3x)\sec^2(3x)\,dx = \int u \cdot \frac{1}{3}\,du = \frac{1}{3}\cdot\frac{u^2}{2} = \frac{1}{6}\tan^2(3x). \]

So the right-hand integral becomes \(-\frac{1}{6}\tan^2(3x)\), and the implicit general solution is

\[ \boxed{y^2 = -\frac{1}{6}\tan^2(3x)+C.} \]

(You can double-check this by differentiating implicitly: \(2yy' = -\frac{1}{6}\cdot 2\tan(3x)\sec^2(3x)\cdot 3 = -\tan(3x)\sec^2(3x)\), which matches the original equation once you divide both sides by \(2y\).)

Solution:

Separate variables by dividing both sides by \(y\ln(y)\):

\[ \frac{dy}{y\ln(y)}=-dx. \]

For the left side, use the substitution \(w=\ln(y)\), so \(dw=\frac{1}{y}dy\), turning the integral into \(\int \frac{dw}{w}\):

\[ \int \frac{1}{y\ln(y)}dy=\int\frac{dw}{w} = \ln|w| = \ln|\ln(y)|. \]

So integrating both sides of the differential equation gives

\[ \ln|\ln(y)|=-x+C. \]

Exponentiate both sides to remove the outer logarithm (absorbing the sign and the constant of exponentiation into a single new constant \(C\)):

\[ \ln(y)=Ce^{-x}. \]

Apply the initial condition \(y(0)=e\), i.e. \(\ln(y(0)) = \ln(e) = 1\):

\[ 1=Ce^{0} \Rightarrow C=1. \]

So \(\ln(y)=e^{-x}\), and exponentiating one more time gives the particular solution:

\[ \boxed{y=e^{e^{-x}}.} \]

Solution:

First, combine the exponentials on the right-hand side, since they share a base:

\[ e^x y \frac{dy}{dx} = e^{-2x-2y}. \]

To separate variables, we want all the \(x\)-dependence on one side. Multiply both sides by \(e^{-x}\):

\[ y\frac{dy}{dx} = e^{-x}\cdot e^{-2x-2y} = e^{-3x-2y}. \]

Now split the exponential \(e^{-3x-2y} = e^{-3x}e^{-2y}\), and move the \(e^{-2y}\) factor to the left side (dividing) so that all \(y\)'s are on the left and all \(x\)'s are on the right:

\[ y\, e^{2y}\, dy = e^{-3x}\, dx. \]

Integrate both sides:

\[ \int y e^{2y}\, dy = \int e^{-3x}\, dx. \]

The right side is immediate: \(\int e^{-3x}dx = -\frac{1}{3}e^{-3x}+C\). The left side needs integration by parts, with \(u=y\), \(dv=e^{2y}dy\) (so \(du=dy\), \(v=\frac{1}{2}e^{2y}\)):

\[ \int y e^{2y}\,dy = \frac{1}{2}ye^{2y} - \int \frac{1}{2}e^{2y}\,dy = \frac{1}{2}ye^{2y}-\frac{1}{4}e^{2y}. \]

Putting both sides together, the implicit general solution is

\[ \boxed{\frac{1}{2}ye^{2y}-\frac{1}{4}e^{2y} = -\frac{1}{3}e^{-3x}+C.} \]

Solution:

Separating variables gives \(\frac{dy}{y}=\frac{dx}{x}\), which integrates to \(\ln|y|=\ln|x|+C\), i.e. \(y=Cx\) for an arbitrary constant \(C\). But every line \(y=Cx\), no matter what \(C\) is, passes through the origin \((0,0)\) — so infinitely many solution curves all pass through that one point, meaning there cannot be a unique solution there.

The underlying issue is that \(f(x,y)=y/x\) is not even defined at \(x=0\) (division by zero), so \(f\) fails to be continuous at the origin. Since the existence-uniqueness theorem's hypotheses (continuity of \(f\) and \(\partial f/\partial y\) in a rectangle around the point) fail right at \((0,0)\), the theorem simply gives no guarantee there — and, as we just saw, uniqueness genuinely does fail. (Separately, the separation step itself divides by \(x\), which is exactly the operation that breaks down at \(x=0\).)

Solution:

Divide by \(3\) to put the equation in standard form \(y'+P(x)y=f(x)\):

\[ y'+4y=\frac{4}{3}. \]

The integrating factor is \(\mu(x)=e^{\int 4\,dx}\):

\[ \mu(x)=e^{4x}. \]

Multiplying the standard-form equation by \(\mu(x)\) collapses the left side into the derivative of a product:

\[ \frac{d}{dx}(e^{4x}y)=\frac{4}{3}e^{4x}. \]

Integrate both sides:

\[ e^{4x}y=\frac{4}{3}\cdot\frac{1}{4}e^{4x}+C=\frac{1}{3}e^{4x}+C. \]

Divide by \(e^{4x}\) to solve for \(y\):

\[ \boxed{y=\frac{1}{3}+Ce^{-4x}.} \]

Solution:

This is already in standard form, with integrating factor

\[ \mu(x)=e^{\int 3\,dx}=e^{3x}. \]

Multiplying through:

\[ \frac{d}{dx}(e^{3x}y)=6e^{3x}. \]

Integrating both sides:

\[ e^{3x}y=2e^{3x}+C. \]

Dividing by \(e^{3x}\):

\[ \boxed{y=2+Ce^{-3x}.} \]

Solution:

Move the \(y\)-term to the left to reach standard form:

\[ y'-2y=x^2+5. \]

The integrating factor is

\[ \mu(x)=e^{\int -2\,dx}=e^{-2x}. \]

Multiplying through:

\[ \frac{d}{dx}(e^{-2x}y)=e^{-2x}(x^2+5). \]

The right side needs integration by parts twice (or the tabular/reduction-formula shortcut) to handle \(x^2e^{-2x}\). Using \(\int x^2 e^{-2x}\,dx = e^{-2x}\left(-\frac{x^2}{2}-\frac{x}{2}-\frac{1}{4}\right)+C\) together with \(\int 5e^{-2x}\,dx = -\frac{5}{2}e^{-2x}+C\):

\[ \int e^{-2x}(x^2+5)\,dx = e^{-2x}\left(-\frac{x^2}{2}-\frac{x}{2}-\frac{1}{4}-\frac{5}{2}\right)+C = e^{-2x}\left(-\frac{x^2}{2}-\frac{x}{2}-\frac{11}{4}\right)+C. \]

So

\[ e^{-2x}y = e^{-2x}\left(-\frac{x^2}{2}-\frac{x}{2}-\frac{11}{4}\right)+C, \]

and dividing by \(e^{-2x}\) gives

\[ \boxed{y=-\frac{x^2}{2}-\frac{x}{2}-\frac{11}{4}+Ce^{2x}.} \]

(You can check this answer by plugging it back into \(y'-2y=x^2+5\) directly — the \(Ce^{2x}\) terms should cancel completely, leaving exactly \(x^2+5\).)

Solution:

The equation is already in standard form with \(P(x)=-\tan(x)\), so the integrating factor is

\[ \mu(x)=e^{\int -\tan(x)\,dx}=e^{\ln|\cos x|}=\cos(x). \]

Multiplying through by \(\cos(x)\):

\[ \frac{d}{dx}(y\cos x)=\cos(x)\cdot\cos^2(x) = \cos^3 x. \]

To integrate \(\cos^3x\), split off one cosine and use \(\cos^2x=1-\sin^2x\):

\[ \int\cos^3x\,dx = \int\cos x(1-\sin^2x)\,dx = \sin x-\frac{\sin^3x}{3}+C. \]

So

\[ y\cos x=\sin x-\frac{\sin^3x}{3}+C. \]

Apply the initial condition \(y(0)=-1\): at \(x=0\), \(\cos(0)=1\) and \(\sin(0)=0\), so

\[ (-1)(1) = 0-0+C \quad\Longrightarrow\quad C=-1. \]

So \(y\cos x = \sin x - \frac{1}{3}\sin^3x-1\), and dividing by \(\cos x\) gives the particular solution:

\[ \boxed{y=\frac{\sin x-\frac{1}{3}\sin^3x-1}{\cos x}.} \]

Solution:

This equation is not linear in \(y\), but it becomes linear if we substitute \(u=y^2\). First, rewrite the equation as \[ yy'-2y^2=x. \]

Let \(u=y^2\). Then \(\dfrac{du}{dx}=2y\dfrac{dy}{dx}=2yy'\), so \(yy'=\dfrac{1}{2}\dfrac{du}{dx}\). Substituting,

\[ \frac{1}{2}\frac{du}{dx}-2u=x \quad\Longrightarrow\quad \frac{du}{dx}-4u=2x. \]

This is now a first-order linear equation in \(u\). The integrating factor is \(\mu(x)=e^{-4x}\), and

\[ \frac{d}{dx}(e^{-4x}u)=2xe^{-4x}. \]

Integrating the right side (using \(\int xe^{ax}dx = e^{ax}\left(\frac{x}{a}-\frac{1}{a^2}\right)\) with \(a=-4\)):

\[ \int 2xe^{-4x}\,dx = e^{-4x}\left(-\frac{x}{2}-\frac{1}{8}\right)+C. \]

So

\[ u = Ce^{4x}-\frac{x}{2}-\frac{1}{8}, \]

and recalling \(u=y^2\),

\[ y^2=Ce^{4x}-\frac{x}{2}-\frac{1}{8}. \]

Now apply the initial condition \(y(1)=5\), i.e. \(y^2=25\) when \(x=1\):

\[ 25=Ce^4-\frac{1}{2}-\frac{1}{8} = Ce^4-\frac{5}{8} \quad\Longrightarrow\quad Ce^4 = 25+\frac{5}{8}=\frac{205}{8} \quad\Longrightarrow\quad C=\frac{205}{8}e^{-4}. \]

So \(Ce^{4x} = \frac{205}{8}e^{-4}e^{4x} = \frac{205}{8}e^{4(x-1)}\), and the particular solution is

\[ \boxed{y^2=\frac{205}{8}e^{4(x-1)}-\frac{x}{2}-\frac{1}{8}.} \]

Solution:

The entire method hinges on one algebraic trick: multiplying the standard-form equation by \(\mu(x)\) turns the left side into the exact derivative of a product, \[ \mu(x)y' + \mu(x)P(x)y = \frac{d}{dx}\big(\mu(x)y\big), \] which is possible only because \(y\) and \(y'\) each appear to the first power (linearly) and are not multiplied by each other — that is exactly what makes the product rule \(\frac{d}{dx}(\mu y) = \mu y' + \mu' y\) line up with the left side of the equation once \(\mu\) is chosen correctly.

If we try the same idea on \(y'+y^2=x\), multiplying by some \(\mu(x)\) gives \(\mu(x)y' + \mu(x)y^2\). For this to be a total derivative of some product, we would need something like \(\frac{d}{dx}(\mu y) = \mu y' + \mu' y\) — but our second term is \(\mu(x)y^2\), not \(\mu'(x)y\). There is no function \(\mu(x)\) depending on \(x\) alone that can turn \(\mu(x)y^2\) into a matching term, because the derivative of any product involving \(y^2\) naturally produces a \(2yy'\) term (via the chain rule), which is not present in our expression at all. So the product-rule trick that makes the whole method work is fundamentally tied to \(y\) appearing linearly — it breaks down as soon as \(y\) is raised to a power other than one.

Solution:

Identify \(M\) and \(N\) from \(M\,dx+N\,dy=0\):

\[ M=2x+y, \qquad N=-(x+6y). \]

The equation is exact exactly when \(M_y=N_x\), so compute both partial derivatives:

\[ M_y=\frac{\partial}{\partial y}(2x+y)=1, \qquad N_x=\frac{\partial}{\partial x}\big(-(x+6y)\big)=-1. \]

Since \(M_y=1\neq -1=N_x\), the equation is not exact, and the method of this section does not apply to it directly (it would require an integrating factor, which is beyond this section).

Solution:

Here \(M=2xy-3\) and \(N=x^2+4y\). Check exactness:

\[ M_y=2x, \qquad N_x=2x. \]

Since \(M_y=N_x\), the equation is exact.

Integrate \(M\) with respect to \(x\):

\[ f=\int(2xy-3)\,dx = x^2y-3x+g(y). \]

Differentiate with respect to \(y\) and match to \(N\):

\[ \frac{\partial f}{\partial y}=x^2+g'(y)\stackrel{\text{set}}{=}x^2+4y \quad\Longrightarrow\quad g'(y)=4y \quad\Longrightarrow\quad g(y)=2y^2. \]

The implicit general solution is

\[ \boxed{x^2y-3x+2y^2=C.} \]

Solution:

Rewrite the equation in the standard order \(M\,dx+N\,dy=0\):

\[ M=-2y, \qquad N=5y-2x. \]

Check exactness:

\[ M_y=-2, \qquad N_x=-2. \]

Since \(M_y=N_x\), the equation is exact.

To find \(f(x,y)\) with \(f_x=M\), integrate \(M\) with respect to \(x\) (treating \(y\) as a constant), which introduces an unknown function \(g(y)\) instead of a constant:

\[ f=\int -2y\,dx=-2xy+g(y). \]

Differentiate this with respect to \(y\) and match it to \(N\):

\[ \frac{\partial f}{\partial y}=-2x+g'(y) \stackrel{\text{set}}{=} 5y-2x \quad\Longrightarrow\quad g'(y)=5y. \]

Integrate to find \(g(y)\):

\[ g(y)=\frac{5}{2}y^2. \]

Assembling \(f(x,y)\), the implicit general solution \(f(x,y)=C\) is

\[ \boxed{-2xy+\frac{5}{2}y^2=C.} \]

Solution:

Here \(M=6xy^3+\cos y\) and \(N=2kx^2y^2-x\sin y\) involve the unknown constant \(k\); we need to choose \(k\) to force \(M_y=N_x\). Compute both partial derivatives:

\[ M_y=18xy^2-\sin y, \qquad N_x=4kxy^2-\sin y. \]

Setting \(M_y=N_x\), the \(-\sin y\) terms cancel from both sides, leaving

\[ 18xy^2=4kxy^2. \]

Since this must hold for all \(x\) and \(y\) (not just specific values), we can divide both sides by \(xy^2\) (assuming it's nonzero) to compare coefficients directly:

\[ 18=4k \quad\Longrightarrow\quad k=\frac{9}{2}. \]

With this value of \(k\), the equation is exact for every \(x\) and \(y\), and could then be solved by the usual method (integrate \(M\) in \(x\), match \(\partial f/\partial y\) to \(N\), and solve for the unknown function of \(y\)).

Solution:

No — \(M_y\neq N_x\) only tells us that \(M\,dx+N\,dy\) is not already an exact differential; it does not mean the equation has no solution at all, only that the specific method of this section (integrate \(M\), match to \(N\)) does not apply directly.

The standard rescue is to multiply the whole equation by a well-chosen integrating factor \(\mu(x)\) or \(\mu(y)\) (a function of \(x\) alone, or of \(y\) alone, depending on the equation), turning \(\mu M\,dx+\mu N\,dy=0\) into an exact equation even though the original one wasn't. This is a different (and more general) use of the phrase "integrating factor" than the one from Section 2.3, though the underlying idea — multiplying by a function to make a derivative pattern appear — is philosophically the same trick.

Solution:

Step 1 (Substitute).

The right side depends only on the combination \(x+y-1\), so let \(u=x+y-1\). Then \(\dfrac{du}{dx}=1+\dfrac{dy}{dx}\), i.e. \(\dfrac{dy}{dx}=\dfrac{du}{dx}-1\).

Step 2 (Rewrite the DE in terms of \(u\)).

\[ \frac{du}{dx}-1=u^2 \quad\Longrightarrow\quad \frac{du}{dx}=u^2+1. \]

Step 3 (Separate and integrate).

\[ \frac{du}{u^2+1}=dx \quad\Longrightarrow\quad \int\frac{du}{u^2+1}=\int dx \quad\Longrightarrow\quad \tan^{-1}(u)=x+C. \]

Step 4 (Return to the original variables).

Solving for \(u\), \(u=\tan(x+C)\). Since \(u=x+y-1\), this gives \(x+y-1=\tan(x+C)\), so

\[ \boxed{y=\tan(x+C)-x+1.} \]

Solution:

Step 1 (Substitute).

Let \(u=y-x+2\), so \(\dfrac{du}{dx}=\dfrac{dy}{dx}-1\), i.e. \(\dfrac{dy}{dx}=\dfrac{du}{dx}+1\).

Step 2 (Rewrite the DE in terms of \(u\)).

\[ \frac{du}{dx}+1=1+\sqrt{u} \quad\Longrightarrow\quad \frac{du}{dx}=\sqrt{u}. \]

Step 3 (Separate and integrate).

\[ \frac{du}{\sqrt{u}}=dx \quad\Longrightarrow\quad \int u^{-1/2}\,du=\int dx \quad\Longrightarrow\quad 2\sqrt{u}=x+C. \]

Solving for \(u\):

\[ u=\frac{(x+C)^2}{4}. \]

Step 4 (Return to the original variables).

Since \(u=y-x+2\), we get \(y-x+2=\dfrac{(x+C)^2}{4}\), so

\[ \boxed{y=x-2+\frac{(x+C)^2}{4}.} \]

Solution:

This one is harder because the substitution isn't handed to you — you have to notice that \(x^2+2xy+y^2\) is a perfect square: \[ x^2+2xy+y^2=(x+y)^2, \] so the equation is really \[ \frac{dy}{dx}=(x+y)^2-1, \] which now has the familiar form \(f(Ax+By+C)\) with \(A=B=1\), \(C=0\).

Step 1 (Substitute).

Let \(u=x+y\), so \(\dfrac{du}{dx}=1+\dfrac{dy}{dx}\), i.e. \(\dfrac{dy}{dx}=\dfrac{du}{dx}-1\).

Step 2 (Rewrite the DE in terms of \(u\)).

\[ \frac{du}{dx}-1=u^2-1 \quad\Longrightarrow\quad \frac{du}{dx}=u^2. \]

Step 3 (Separate and integrate).

\[ \frac{du}{u^2}=dx \quad\Longrightarrow\quad \int u^{-2}\,du=\int dx \quad\Longrightarrow\quad -\frac{1}{u}=x+C. \]

Solving for \(u\):

\[ u=-\frac{1}{x+C}. \]

Step 4 (Return to the original variables).

Since \(u=x+y\), we get \(x+y=-\dfrac{1}{x+C}\), so

\[ \boxed{y=-x-\frac{1}{x+C}.} \]

Solution:

Differentiating \(u=Ax+By+C\) with respect to \(x\) gives \(\dfrac{du}{dx}=A+B\dfrac{dy}{dx}\). As long as \(B\neq 0\), we can solve this for \(\dfrac{dy}{dx}\): \[ \frac{dy}{dx}=\frac{1}{B}\left(\frac{du}{dx}-A\right). \] Substituting this into the original equation replaces \(\dfrac{dy}{dx}\) with an expression purely in \(\dfrac{du}{dx}\), and the right-hand side \(f(Ax+By+C)\) is already just \(f(u)\) — so the whole equation becomes a relationship between \(\dfrac{du}{dx}\), \(u\), and \(x\) only, with no \(y\) left anywhere. That is precisely what "separable in \(u\) and \(x\)" means, and it is always achievable this way because \(B\neq 0\) let us isolate \(dy/dx\).

If \(B=0\), the substitution \(u=Ax+C\) would not involve \(y\) at all, so there would be no way to recover \(\dfrac{dy}{dx}\) from \(\dfrac{du}{dx}\) — the substitution loses all information about \(y\) and becomes useless. But this case is also not a problem in practice: if \(B=0\), the original equation is \(\dfrac{dy}{dx}=f(Ax+C)\), whose right-hand side already depends on \(x\) alone. Such an equation is already directly integrable (\(y=\int f(Ax+C)\,dx\)) without needing any substitution trick at all.