Practice / Review Problems — MATH 2410

Solution:

Compute \(y'\): \[ y' = \frac{d}{dx} e^{-x/2} = -\frac{1}{2} e^{-x/2} \] Substitute into the DE: \[ 2y' + y = 2\left(-\frac{1}{2} e^{-x/2}\right) + e^{-x/2} = 0 \] So \(y = e^{-x/2}\) is a solution on \((-\infty, \infty)\).

Solution:

Differentiate implicitly: \[ \frac{d}{dx}(x^2 + y^2) = 0 \implies 2x + 2y \frac{dy}{dx} = 0 \] Solve for \(y'\): \[ \frac{dy}{dx} = -\frac{x}{y} \] So the family satisfies the DE. The interval excludes points where \(y=0\).

Solution:

Compute \(y'\): \[ y' = \frac{1}{2\sqrt{x}} \int_4^x \frac{\cos(t)}{\sqrt{t}} dt + \cos(x) \] Multiply by \(2x\): \[ 2x y' = 2x \cos(x) + \sqrt{x} \int_4^x \frac{\cos(t)}{\sqrt{t}} dt = 2x \cos(x) + y \] Hence, \[ 2xy' - y = 2x \cos(x) \]

Solution:

• Type: Ordinary differential equation, since only ordinary derivatives of \(y\) appear.
• Order: Second order, because the highest derivative is \(y''\).
• Linearity: Linear, since \(y\) and its derivatives appear only to the first power and are not multiplied together.

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.

Solution:

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

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.

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 \(x = \pm 1\). Since the initial condition occurs at \(x = 2\), the largest interval containing \(2\) 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, both \(f(x,y)\) and \(\frac{\partial f}{\partial y}\) must be continuous in a region containing the point \((x_0, y_0)\).

First, observe that \(\sqrt{xy}\) is defined only 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 is undefined when \(xy = 0\) and not real when \(xy < 0\). Therefore, both \(f(x,y)\) and \(\frac{\partial f}{\partial y}\) are continuous only when \[ xy > 0. \]

One such region is \[ x > 0 \quad \text{and} \quad y > 0. \]

In this region, the 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:

Let \(P(t)\) denote the population at time \(t\).

The birth rate is proportional to the population, so we model it as \[ \text{Birth rate} = aP, \] where \(a > 0\) is a 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 constant.

The net rate of change of the population is given by \[ \frac{dP}{dt} = (\text{birth rate}) - (\text{death rate}). \]

Substituting the expressions above, we obtain the differential equation \[ \frac{dP}{dt} = aP - bP^2. \]

This is a first-order, nonlinear differential equation known as the logistic equation.

Solution:

Let \(A(t)\) be the amount of salt (in pounds) in the tank at time \(t\).

The rate of change of salt in the tank is given by \[ \frac{dA}{dt} = (\text{rate in}) - (\text{rate out}). \]

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

Rate out:
The volume of liquid in the tank is increasing because the inflow exceeds the outflow: \[ V(t) = 300 + (3 - 2)t = 300 + t \text{ gal}. \]

The concentration of salt in the tank at time \(t\) is \[ \frac{A(t)}{300 + t} \text{ lb/gal}. \]

Since the solution is pumped out at \(2\) gal/min, the rate of salt leaving the tank is \[ \text{Rate out} = 2 \cdot \frac{A(t)}{300 + t}. \]

Therefore, the differential equation is \[ \frac{dA}{dt} = 6 - \frac{2A}{300 + t}. \]

Solution:

Let \(x(t)\) denote the amount of drug in the bloodstream at time \(t\).

The rate of change of the drug amount is given by \[ \frac{dx}{dt} = (\text{rate in}) - (\text{rate out}). \]

The drug is infused at a constant rate \(r\) grams per second, 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, we obtain \[ \frac{dx}{dt} = r - kx. \]

This is a first-order linear differential equation commonly used in pharmacokinetics.

Solution:

First, factor the right-hand side: \[ y' = y^2(1 - y). \]

Critical points occur when \( y' = 0 \): \[ y = 0, \quad y = 1. \]

To analyze stability, examine the sign of \(y'\):

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

Classification:

• \(y=0\): semi-stable
• \(y=1\): asymptotically stable

Solution:

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

Critical points occur when \(y' = 0\): \[ y = 0, \quad y = 2, \quad y = 4. \]

Sign analysis:

• \(y < 0\): \(y' < 0\)
• \(0 < y < 2\): \(y' > 0\)
• \(2 < y < 4\): \(y' < 0\)
• \(y > 4\): \(y' > 0\)

Classification:

• \(y=0\): unstable
• \(y=2\): asymptotically stable
• \(y=4\): unstable

Solution:

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

Since \(e^y > 0\) for all \(y\), the sign of \(y'\) depends only on \(y(e^y - 9)\).

Critical points occur when: \[ y = 0, \quad e^y = 9 \Rightarrow y = \ln 9. \]

Sign analysis:

• \(y < 0\): \(y' > 0\)
• \(0 < y < \ln 9\): \(y' < 0\)
• \(y > \ln 9\): \(y' > 0\)

Classification:

• \(y=0\): asymptotically stable
• \(y=\ln 9\): unstable

Solution:

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

If a nonconstant solution \(y(x)\) were to cross \(y=c\), then at the crossing point both solutions would have the same value.

However, the existence and uniqueness theorem guarantees that a first-order autonomous differential equation has a unique solution through any initial point \((x_0, y_0)\).

Therefore, once a solution reaches \(y=c\), it must remain there for all \(x\), and crossing is impossible.

This is why solution curves of autonomous equations never cross equilibrium solutions.

Solution:

Rewrite the equation:

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

Multiply both sides by \(e^{-x}\):

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

Separate variables:

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

Integrate both sides:

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

Use integration by parts on the left:

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

Solution:

Separate variables:

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

Simplify:

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

Integrate:

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

Let \(u=\tan(3x)\):

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

Solution:

Separate variables:

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

Integrate:

\[ \int \frac{1}{y\ln(y)}dy=-\int dx \] \[ \ln|\ln(y)|=-x+C \]

Exponentiate:

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

Apply initial condition:

\[ \ln(e)=C \Rightarrow C=1 \]

Final Answer:

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

Solution:

Standard form:

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

Integrating factor:

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

Multiply equation:

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

Integrate:

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

Final Answer:

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

Solution:

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

Integrate:

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

Solution:

Rewrite:

\[ y' -2y=\frac{x}{y} \]

Multiply by \(y\):

\[ y y' -2y^2=x \]

Let \(u=y^2\):

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

Use initial condition:

\[ 25=C e^4-\frac{1}{2}-\frac{1}{8} \]

Solution:

\[ \mu=e^{\int -\tan(x)dx}=\cos(x) \] \[ \frac{d}{dx}(y\cos x)=\cos^3 x \] \[ y\cos x=\int \cos^3 x dx \] \[ y\cos x=\sin x-\frac{\sin^3 x}{3}+C \]

Apply initial condition:

\[ C=-1 \]

Solution:

\[ M=2x+y, \quad N=-(x+6y) \] \[ M_y=1, \quad N_x=-1 \]

Not exact.

Solution:

\[ M=-2y, \quad N=5y-2x \] \[ M_y=-2, \quad N_x=-2 \]

Exact.

\[ f=\int -2y dx=-2xy+g(y) \] \[ \frac{\partial f}{\partial y}=-2x+g'(y)=5y-2x \] \[ g'(y)=5y \Rightarrow g=\frac{5}{2}y^2 \] \[ -2xy+\frac{5}{2}y^2=C \]

Solution:

\[ M_y=18xy^2-\sin y \] \[ N_x=4kxy^2-\sin y \] \[ 18xy^2=4kxy^2 \Rightarrow k=\frac{9}{2} \]