Chapter 7: Differential Equations

Solution. Integrate both sides with respect to $x$:

Apply the initial condition $y(1) = 3$:

So $C = 2$, and the particular solution is $y = 2x^3 - 2x^2 + x + 2$.

Solution. Differentiate: $\frac{dy}{dx} = 3Ce^{3x} = 3y$. This confirms the equation is satisfied for every value of $C$.

For the particular solution, apply $y(0) = 7$:

Therefore $C = 7$ and $y = 7e^{3x}$.

Solution. Compute $f(x, y) = x - y$ at each grid point:

At $(-1,-1)$, $(0,0)$, and $(1,1)$, draw horizontal segments (slope $= 0$). At $(1,-1)$, draw a steeply rising segment (slope $= 2$). At $(-1,1)$, draw a steeply falling segment (slope $= -2$). The segments along the line $y = x$ are all horizontal, confirming that $y = x$ is related to the equilibrium behavior of this DE.

(A) $\frac{dy}{dx} = -y$   (B) $\frac{dy}{dx} = -x$   (C) $\frac{dy}{dx} = x - y$   (D) $\frac{dy}{dx} = y^2$

Solution. Horizontal segments along the $x$-axis mean $\frac{dy}{dx} = 0$ when $y = 0$. This eliminates (B), since $-x \neq 0$ for all $x$, and (D), since $y^2 = 0$ when $y = 0$ but $y^2 \geq 0$ always, so segments below the $x$-axis would also point upward or be horizontal. For (C), at $(1, 0)$ the slope is $1$, which is positive, not zero—so slopes along the $x$-axis are not all zero. Only (A) satisfies all conditions: $\frac{dy}{dx} = -y$ gives slope $0$ on the $x$-axis, negative slopes when $y > 0$, and positive slopes when $y < 0$. The answer is (A).

Solution. First, identify equilibrium solutions by setting $\frac{dy}{dx} = 0$:

These are two horizontal equilibrium lines. Note this DE depends only on $y$ (not $x$), so all segments in the same horizontal row have the same slope.

• When $0 < y < 2$: $y > 0$ and $(2-y) > 0$, so $\frac{dy}{dx} > 0$ (slopes point upward).
• When $y > 2$: $y > 0$ but $(2-y) < 0$, so $\frac{dy}{dx} < 0$ (slopes point downward).
• When $y < 0$: $y < 0$ and $(2-y) > 0$, so $\frac{dy}{dx} < 0$ (slopes point downward).

Starting at $(0, 1)$: since $0 < 1 < 2$, the slope is positive and the solution increases. As $y$ approaches 2, $\frac{dy}{dx} \to 0$ and the curve levels off. The solution curve approaches $y = 2$ asymptotically from below. The equilibrium $y = 2$ is stable (solutions near it converge to it), while $y = 0$ is unstable (solutions near it diverge from it).

Solution. Equilibrium: $\frac{dy}{dx} = 0$ when $(y-1)(y-4) = 0$, so $y = 1$ and $y = 4$.

Analyze the sign of $\frac{dy}{dx}$ in each region:

• $y < 1$: $(y-1) < 0$, $(y-4) < 0$, product is $> 0$. Solutions increase toward $y = 1$.
• $1 < y < 4$: $(y-1) > 0$, $(y-4) < 0$, product is $< 0$. Solutions decrease toward $y = 1$.
• $y > 4$: $(y-1) > 0$, $(y-4) > 0$, product is $> 0$. Solutions increase away from $y = 4$.

Solutions above and below $y = 1$ both move toward it, so $y = 1$ is stable.

Solutions just below $y = 4$ move toward $y = 1$ (away from $y = 4$), and solutions just above $y = 4$ increase further away. So $y = 4$ is unstable.

• All segments along $y = 3$ are horizontal.
• Segments where $y > 3$ have negative slopes.
• Segments where $y < 3$ have positive slopes.
• The steepness increases as $y$ moves farther from 3.

Which differential equation matches? (A) $\frac{dy}{dx} = 3 - y$   (B) $\frac{dy}{dx} = y - 3$   (C) $\frac{dy}{dx} = (3 - y)^2$   (D) $\frac{dy}{dx} = 3 - x$

Solution.

Eliminate (D): slopes depend on $x$, not $y$, so segments in the same column (not row) would have equal slopes—contradicting the described behavior.

Eliminate (C): $(3-y)^2 \geq 0$ always, so slopes can never be negative. But we need negative slopes when $y > 3$.

Test (A) $3 - y$: when $y = 3$, slope $= 0$ (horizontal). When $y > 3$, slope $< 0$ (negative). When $y < 3$, slope $> 0$ (positive). As $|y - 3|$ increases, $|3 - y|$ increases, making slopes steeper. All properties match.

Test (B) $y - 3$: when $y > 3$, slope $> 0$ (positive). This contradicts "segments where $y > 3$ have negative slopes."

The answer is (A).

Solution. We need to step from $x = 0$ to $x = 2$ in steps of $0.5$, requiring 4 iterations.

Therefore, $y(2) \approx 7.125$. (The exact solution is $y = 2e^x - x - 1$, giving $y(2) = 2e^2 - 3 \approx 11.78$, illustrating that the large step size introduces significant error.)

Solution.

So $y(1.3) \approx 4.66$. The exact solution is $y = x^2 + 3$, giving $y(1.3) = 1.69 + 3 = 4.69$. With a small step size, Euler's method is quite close. Since $y'' = 2 > 0$ (concave up), our approximation is an underestimate, consistent with the result.

Solution.

Euler's approximation: $y(2) \approx 0.25$. The exact solution is $y = 4e^{-x}$, giving $y(2) = 4e^{-2} \approx 0.541$.

Since $\frac{d^2y}{dx^2} = y > 0$ (the solution curve is concave up), the tangent line lies below the curve, so Euler's method underestimates the true value. Indeed, $0.25 < 0.541$.

Solution. Separate variables:

Integrate both sides:

Apply $y(0) = 3$: $\frac{9}{2} = 0 + C$, so $C = \frac{9}{2}$.

Since $y(0) = 3 > 0$, we take the positive root: $y = \sqrt{\dfrac{2x^3}{3} + 9}$.

Solution. Separate variables:

Integrate:

Exponentiate both sides:

Apply $y(0) = 200$: $200 = Ae^{0} = A$.

Therefore $y = 200e^{-0.05t}$. This represents exponential decay with rate constant $k = -0.05$.

Solution. Separate variables:

Integrate both sides:

Apply $y(1) = -1$: $-\frac{1}{-1} = \ln 1 + 1 + C \implies 1 = 0 + 1 + C$, so $C = 0$.

Solution. The population satisfies $P(t) = 500e^{kt}$. The doubling condition gives:

After 8 hours:

The colony has approximately 3175 bacteria after 8 hours.

Solution. From the half-life: $\lambda = \frac{\ln 2}{12}$, so $A(t) = 80e^{-\frac{\ln 2}{12}t}$.

Equivalently, $A(t) = 80 \cdot 2^{-t/12}$. After 20 years:

Approximately 25.2 grams remain after 20 years. Notice that after 12 years we'd have 40 g, and after 24 years we'd have 20 g, so 25.2 g at 20 years is reasonable.

(a) What is the carrying capacity?

(b) At what population is the growth rate greatest?

(c) Find $P(t)$.

Solution.

(a) Comparing with the standard form, $L = 5000$. The lake can sustain at most 5000 fish.

(b) The growth rate is maximized when $P = L/2 = 2500$ fish.

(c) We have $k = 0.4$, $L = 5000$, $P_0 = 800$. Compute $A$:

Therefore:

As $t \to \infty$, $e^{-0.4t} \to 0$, so $P(t) \to 5000$, confirming the population approaches the carrying capacity.

(a) $\dfrac{dy}{dx} + 4xy = e^x$   (b) $\dfrac{dy}{dx} = y^2 - x$   (c) $\dfrac{dy}{dx} + y\cos x = 0$   (d) $y\,\dfrac{dy}{dx} = x^2$

Solution.

(a) The equation is already in standard form with $P(x) = 4x$ and $Q(x) = e^x$. $y$ appears only to the first power. LINEAR

(b) The right side contains $y^2$ — $y$ is squared. This violates the first test. NONLINEAR

(c) Rewrite: $\dfrac{dy}{dx} + (\cos x)\,y = 0$. This has $P(x) = \cos x$ and $Q(x) = 0$. The $\cos x$ is a coefficient of $y$, not an argument; $y$ appears to the first power. LINEAR

(d) The left side has $y \cdot \dfrac{dy}{dx}$ — a product of $y$ with its own derivative. This violates the third test. NONLINEAR

Solution. Rewrite: $\dfrac{dy}{dx} - e^x y = 0$. Now $P(x) = -e^x$ and $Q(x) = 0$. The exponential $e^x$ is a coefficient of $y$, and $y$ appears to the first power. LINEAR

Compare this carefully with $\dfrac{dy}{dx} = e^y$, where $y$ is in the exponent. That equation is NONLINEAR because $y$ appears inside an exponential function.

Solution. Divide every term by $(x^2 + 1)$: $$\frac{dy}{dx} - \frac{3x}{x^2+1}\,y = \frac{x}{x^2+1}.$$ This is standard form with $P(x) = -\dfrac{3x}{x^2+1}$ and $Q(x) = \dfrac{x}{x^2+1}$. Although the coefficients are complicated, $y$ appears only to the first power. LINEAR

Solution. Try to rearrange into standard form: $$\frac{dy}{dx} + y = xy^2.$$ The right side is $xy^2$, which contains $y^2$. There is no algebraic manipulation that removes $y^2$. NONLINEAR

This equation is a Bernoulli equation — a special nonlinear type solvable by a substitution $v = y^{1-n}$. You will encounter this in a university ODE course.

Solution. Rewrite: $\dfrac{dy}{dx} - \sin(x)\,y = -x^2$. Standard form with $P(x) = -\sin x$ and $Q(x) = -x^2$. $y$ appears once, to the first power, as part of the coefficient term $-\sin(x)\cdot y$. LINEAR

Solution. Divide by $\sqrt{x}$: $$\frac{dy}{dx} + \frac{1}{x}\,y = \sqrt{x}.$$ Standard form with $P(x) = \dfrac{1}{x}$ (defined for $x > 0$) and $Q(x) = \sqrt{x}$. The complicated-looking $\sqrt{x}$ factors are all functions of $x$ only. LINEAR

(a) $\dfrac{dy}{dx} + 3y = 0$   (b) $\dfrac{dy}{dx} - 2y = e^x$   (c) $\dfrac{dy}{dx} + \dfrac{y}{x} = \ln x$   (d) $\dfrac{dy}{dx} - ky = 0$

Solution.

(a) $Q(x) = 0$. HOMOGENEOUS. General solution: $y = Ce^{-3x}$.

(b) $Q(x) = e^x \neq 0$. NONHOMOGENEOUS. Requires an integrating factor or undetermined coefficients to solve.

(c) $Q(x) = \ln x \neq 0$ (for $x > 0$). NONHOMOGENEOUS.

(d) $Q(x) = 0$. HOMOGENEOUS. This is exactly the exponential growth/decay equation — its solution $y = Ce^{kx}$ is already familiar from Section 7.5.

Solution. First check linearity. Rewrite:

Divide by $x$ (assuming $x \neq 0$):

$P(x) = \dfrac{2}{x}$, $Q(x) = 5 \neq 0$. The equation is linear and NONHOMOGENEOUS.

(a) $\dfrac{dy}{dx} = ky$   (b) $\dfrac{dy}{dx} = ky + b$   (c) $\dfrac{dy}{dx} = \dfrac{L - y}{L}\cdot y$   (d) $\dfrac{dy}{dx} + \dfrac{y}{x} = x\sin x$

Solution.

(a) Rewrite: $\dfrac{dy}{dx} - ky = 0$. $y$ appears to the first power, right side zero. Linear, Homogeneous. This is the exponential growth model from Section 7.5. Solution: $y = Ce^{kx}$.

(b) Rewrite: $\dfrac{dy}{dx} - ky = b$ (where $b$ is a nonzero constant). $y$ is linear; right side $Q = b \neq 0$. Linear, Nonhomogeneous. This models growth with a constant input rate $b$. Separation still works here: $\dfrac{dy}{ky+b} = dx$, giving $y = Ce^{kx} - \dfrac{b}{k}$.

(c) Expand: $\dfrac{dy}{dx} = \dfrac{Ly - y^2}{L} = y - \dfrac{y^2}{L}$. The $y^2$ term makes this Nonlinear. (This is the logistic growth equation from Section 7.6, which is separable despite being nonlinear.)

(d) Already in standard form with $P(x) = \dfrac{1}{x}$ and $Q(x) = x\sin x$. No $y^n$ or $f(y)$ terms. Linear, Nonhomogeneous.

(a) $\dfrac{dy}{dx} = \dfrac{x}{y}$   (b) $\dfrac{dy}{dx} + 2y = 4x$   (c) $\dfrac{dy}{dx} = x + y^2$   (d) $\dfrac{dy}{dx} + 3y = 0$

Solution.

(a) $y\,dy = x\,dx$: the equation is separable (nonlinear but separable). Integrate both sides: $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$, so $y^2 - x^2 = K$. Method: Separation of Variables.

(b) Linear nonhomogeneous. Not directly separable because of the $4x$ term. At AP level, Euler's method estimates solutions numerically. In university, an integrating factor $\mu = e^{2x}$ yields the exact solution $y = 2x - 1 + Ce^{-2x}$. Method: Integrating Factor (university) or Euler's Method (AP).

(c) Nonlinear and non-separable ($x$ and $y^2$ cannot be factored apart). No elementary closed-form solution exists. Method: Euler's Method (numerical approximation) or power series (university).

(d) Linear homogeneous. Also separable: $\dfrac{dy}{y} = -3\,dx \Rightarrow \ln|y| = -3x + C_1 \Rightarrow y = Ce^{-3x}$. Method: Separation of Variables (or direct formula).

Solution. No. To test linearity, we try to rearrange into standard form. Multiply both sides by $(x+y)$:

The left side contains $y\cdot\dfrac{dy}{dx}$, which is a product of $y$ with its derivative — a violation of the third linearity test. The equation is NONLINEAR.

Note: This equation is actually a "degree-0 homogeneous nonlinear DE" (in the second sense of homogeneous). Substituting $v = y/x$ separates it into a solvable form — a technique studied in university ODE courses.

Solution. The general form of a linear second-order ODE is $a(x)y'' + b(x)y' + c(x)y = f(x)$. Here $a = 1$, $b = 3$, $c = -4$, and $f(x) = \sin x$. The dependent variable $y$ and its derivatives $y', y''$ appear only to the first power, with no products like $yy'$ or functions like $\sin(y)$. LINEAR.

Since $f(x) = \sin x \neq 0$, this is NONHOMOGENEOUS. This is a key equation type in university-level ODE courses, solved by finding the complementary solution (of $y'' + 3y' - 4y = 0$) plus a particular solution of the full equation.