Differential Equation

LessonAbout 1.5 h

Differential Equations

Key Ideas

  • A differential equation contains derivatives of an unknown function.
  • General solutions contain arbitrary constants; particular solutions have none.
  • Equations are classified by their order, degree, and type.

Definition: Types of Differential Equations

A differential equation is an equation that contains at least one derivative or differential of an unknown function. Differential equations are categorized into two fundamental types based on the number of independent variables involved:

  • Ordinary DE: Contains only one independent variable and ordinary derivatives.
  • Partial DE: Contains two or more independent variables and partial derivatives.
  • General Solution: Contains one or more arbitrary constants.
  • Particular Solution: Contains no arbitrary constants after applying boundary conditions.

Ordinary Differential Equation Standard Form

\[ \dfrac{dy}{dx} + P(x)y = Q(x) \]

Example. Classify the equations \((2xy + x^2)dx + 2ydy = 0\) and \(\dfrac{\partial x}{\partial y} + 2y\dfrac{\partial x}{\partial z} = 3\) as ordinary or partial differential equations.

Solution. The first equation involves only one independent variable \(x\) and ordinary differentials, so it is an ordinary differential equation. The second equation contains partial derivatives with respect to two independent variables \(y\) and \(z\), making it a partial differential equation.

Answer: First equation is an Ordinary DE; second equation is a Partial DE.

Order of Differential Equations

The order of a differential equation is defined as the order of the highest-ordered derivative that appears in the equation.

Order Criterion

\[ \text{Order} = \text{Order of the highest derivative present} \]

Example. Determine the order of the differential equation \(2x \dfrac{d^4 y}{dx^4} + 5x^2 \left(\dfrac{dy}{dx}\right)^3 - xy = 0\).

Solution. Inspecting the derivatives present in the equation, the derivatives are \(\dfrac{d^4 y}{dx^4}\) (fourth order) and \(\dfrac{dy}{dx}\) (first order). The highest-ordered derivative present is \(\dfrac{d^4 y}{dx^4}\).

Answer: Fourth order.

Degree of Differential Equations

The degree of a differential equation is the exponent or power to which the highest-ordered derivative is raised, after the equation has been cleared of radicals or fractional exponents regarding the derivatives.

Degree Criterion

\[ \text{Degree} = \text{Exponent of the highest-ordered derivative} \]

Example. Determine the degree of the differential equation \(2x \dfrac{d^4 y}{dx^4} + 5x^2 \left(\dfrac{dy}{dx}\right)^3 - xy = 0\).

Solution. The highest-ordered derivative in the equation is \(\dfrac{d^4 y}{dx^4}\). The power or exponent of this term is \(1\). Therefore, the degree of the differential equation is \(1\).

Answer: First degree.

Variable Separable Differential Equations

A first-order differential equation is variable separable if it can be written such that the coefficient of \(dx\) is purely a function of \(x\) and the coefficient of \(dy\) is purely a function of \(y\).

Variable Separable Standard Form & General Solution

\[ P(x)dx + Q(y)dy = 0 \]
\[ \int P(x)dx + \int Q(y)dy = C \]

Example. Solve the differential equation \(x(y - 1)dx + (x + 1)dy = 0\). Given \(y = 2\) when \(x = 1\), find \(y\) when \(x = 2\).

Solution. Separate the variables by dividing throughout by \((x + 1)(y - 1)\):

\[ \dfrac{x}{x + 1}dx + \dfrac{1}{y - 1}dy = 0 \implies \left(1 - \dfrac{1}{x + 1}\right)dx + \dfrac{1}{y - 1}dy = 0 \]

Integrate each term:

\[ x - \ln(x + 1) + \ln(y - 1) = C \]

Substitute boundary condition \(x = 1, y = 2\) to find \(C\):

\[ 1 - \ln(2) + \ln(1) = C \implies C = 1 - \ln 2 \approx 0.307 \]

Substitute \(x = 2\) into the general solution equation:

\[ 2 - \ln(3) + \ln(y - 1) = 0.307 \implies \ln(y - 1) = -0.594 \implies y - 1 = e^{-0.594} \approx 0.552 \]

Answer: \(y = 1.55\)

Homogeneous First-Order Differential Equations

A differential equation \(M(x,y)dx + N(x,y)dy = 0\) is homogeneous if \(M(x,y)\) and \(N(x,y)\) are homogeneous functions of the same degree \(n\), meaning \(f(rx, ry) = r^n f(x,y)\). Replacing \(y = vx\) converts the equation into a variable separable form.

Homogeneous Substitution

\[ y = vx, \qquad\qquad dy = v\,dx + x\,dv \]

Example. Solve the differential equation \((x + y)dy = (x - y)dx\).

Solution. Substitute \(y = vx\) and \(dy = v\,dx + x\,dv\) into the equation:

\[ (x + vx)(v\,dx + x\,dv) = (x - vx)dx \]

Divide by \(x\) and expand terms:

\[ (1 + v)(v\,dx + x\,dv) = (1 - v)dx \implies (v + v^2)dx + x(1 + v)dv = (1 - v)dx \]
\[ (v^2 + 2v - 1)dx + x(1 + v)dv = 0 \]

Separate variables and integrate:

\[ \dfrac{dx}{x} + \dfrac{1 + v}{v^2 + 2v - 1}dv = 0 \implies \ln x + \dfrac{1}{2}\ln(v^2 + 2v - 1) = c \]
\[ \ln[x^2(v^2 + 2v - 1)] = K \implies x^2\left(\dfrac{y^2}{x^2} + \dfrac{2y}{x} - 1\right) = C \]

Answer: \(y^2 + 2xy - x^2 = C\)

Linear First-Order Differential Equations

A first-order differential equation is linear if \(y\) and its derivative \(\dfrac{dy}{dx}\) occur to the first degree and are not multiplied together.

Linear First-Order Standard Form and General Solution

\[ \dfrac{dy}{dx} + P(x)y = Q(x) \]
\[ y \cdot e^{\int P(x)dx} = \int Q(x) e^{\int P(x)dx} dx + C \]

Example. Solve the linear differential equation \(\dfrac{dy}{dx} + \dfrac{y}{x} = x^2\).

Solution. Identify \(P(x) = \dfrac{1}{x}\) and \(Q(x) = x^2\). Compute the integrating factor:

\[ I.F. = e^{\int \frac{1}{x}dx} = e^{\ln x} = x \]

Multiply the differential equation by the integrating factor \(x\) and integrate:

\[ y \cdot x = \int x^2 (x) dx + C = \int x^3 dx + C = \dfrac{x^4}{4} + C \]

Answer: \(xy = \dfrac{x^4}{4} + C\)

Linear Second-Order Differential Equations

A linear second-order differential equation with constant coefficients has the form \(a\dfrac{d^2y}{dx^2} + b\dfrac{dy}{dx} + cy = 0\). Substituting \(D = \dfrac{d}{dx}\) yields the auxiliary equation \(am^2 + bm + c = 0\).

Auxiliary Equation & Solution Cases

\[ am^2 + bm + c = 0 \]
  • Real and distinct roots give exponential sum solutions.
  • Real and equal roots include a linear factor.
  • Complex conjugate roots yield sine and cosine solutions.
\[ y = C_1 e^{m_1 x} + C_2 e^{m_2 x} \qquad\qquad \text{(Real & Distinct)} \]
\[ y = (C_1 x + C_2)e^{m x} \qquad\qquad \text{(Real & Equal)} \]
\[ y = e^{\alpha x}(C_1 \cos \beta x + C_2 \sin \beta x) \qquad\qquad \text{(Complex } m = \alpha \pm i\beta\text{)} \]

Example. Solve the linear second-order differential equation \(2\dfrac{d^2 y}{dx^2} + 5\dfrac{dy}{dx} - 3y = 0\).

Solution. Write the auxiliary equation by replacing derivatives with \(m\):

\[ 2m^2 + 5m - 3 = 0 \implies (2m - 1)(m + 3) = 0 \]

The roots are real and distinct: \(m_1 = \dfrac{1}{2}\) and \(m_2 = -3\). Apply the real and distinct roots general solution formula:

Answer: \(y = C_1 e^{x/2} + C_2 e^{-3x}\)

Exact Differential Equations

A differential equation \(M(x,y)dx + N(x,y)dy = 0\) is exact when the expression represents an exact total differential \(dF(x,y) = M\,dx + N\,dy\).

Exactness Test & Solution Form

\[ \dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} \]
\[ F(x,y) = \int M(x,y)\partial x + k(y) = C \]

Example. Solve \((2xy + 6)dx + (x^2 - 2y - 5)dy = 0\).

Solution. Check for exactness:

\[ M = 2xy + 6 \implies \dfrac{\partial M}{\partial y} = 2x \]
\[ N = x^2 - 2y - 5 \implies \dfrac{\partial N}{\partial x} = 2x \]

Since \(\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}\), the equation is exact. Integrate \(M\) with respect to \(x\) treating \(y\) as constant:

\[ F(x,y) = \int (2xy + 6)\partial x = x^2 y + 6x + k(y) \]

Differentiate \(F\) with respect to \(y\) and equate to \(N\):

\[ \dfrac{\partial F}{\partial y} = x^2 + k'(y) = x^2 - 2y - 5 \implies k'(y) = -2y - 5 \implies k(y) = -y^2 - 5y \]

Answer: \(x^2 y - y^2 + 6x - 5y = C\)

Bernoulli’s Equation

Bernoulli’s equation is a non-linear first-order differential equation of the form \(\dfrac{dy}{dx} + P(x)y = Q(x)y^n\). It is linearized using the substitution \(v = y^{1-n}\).

Bernoulli’s Standard Form & Transformation

\[ \dfrac{dy}{dx} + P(x)y = Q(x)y^n \]
\[ v = y^{1-n} \implies \dfrac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x) \]

Example. Solve the differential equation \(x\dfrac{dy}{dx} + y = 2x^2 y^2\).

Solution. Divide by \(xy^2\) to put into standard form (\(n = 2\)):

\[ y^{-2}\dfrac{dy}{dx} + \dfrac{1}{x}y^{-1} = 2x \]

Let \(v = y^{1-2} = y^{-1}\), then \(\dfrac{dv}{dx} = -y^{-2}\dfrac{dy}{dx}\). Substituting yields:

\[ -\dfrac{dv}{dx} + \dfrac{1}{x}v = 2x \implies \dfrac{dv}{dx} - \dfrac{1}{x}v = -2x \]

This is a linear DE in \(v\) with \(I.F. = e^{\int -1/x dx} = \dfrac{1}{x}\). Multiply by \(I.F.\) and integrate:

\[ v \cdot \dfrac{1}{x} = \int (-2x)\left(\dfrac{1}{x}\right)dx = -2x + C \implies \dfrac{1}{xy} = C - 2x \]

Answer: \(\dfrac{1}{xy} = C - 2x\)

Finding the Differential Equation from a General Solution

To determine the differential equation corresponding to a given general solution, differentiate the general solution as many times as there are independent arbitrary constants, then eliminate the constants among the original and derived equations.

Elimination of Constants Method

\[ \text{Number of Differentiations} = \text{Number of Independent Arbitrary Constants} \]

Example. Find the differential equation whose general solution is \(y^2 = cx\).

Solution. The equation contains one arbitrary constant \(c\). Isolate \(c\):

\[ c = \dfrac{y^2}{x} \]

Differentiate both sides with respect to \(x\) using the quotient rule:

\[ 0 = \dfrac{x(2yy') - y^2(1)}{x^2} \implies 2xy y' - y^2 = 0 \implies y' = \dfrac{y}{2x} \]

Answer: \(2x dy - y dx = 0\)

Applications of DE: Population Growth

Population growth problems assume that the rate of change of a population \(P\) is directly proportional to the population present at any time \(t\).

Population Growth Model

\[ \dfrac{dP}{dt} = kP \implies P(t) = P_0 e^{kt} \]

Example. The population of a country doubles in \(50\text{ years}\). How many years will it take to become \(5\text{ times}\) as much, assuming the rate of increase is proportional to the number of inhabitants?

Solution. Use the exponential growth model \(P(t) = P_0 e^{kt}\). At \(t = 50\), \(P = 2P_0\):

\[ 2P_0 = P_0 e^{50k} \implies 50k = \ln 2 \implies k = \dfrac{\ln 2}{50} \approx 0.013863 \]

Find time \(t\) when \(P = 5P_0\):

\[ 5P_0 = P_0 e^{kt} \implies \ln 5 = kt \implies t = \dfrac{\ln 5}{0.013863} \approx 116.1\text{ years} \]

Answer: \(116\text{ years}\)

Applications of DE: Exponential Growth and Decay

Decay problems model substances (such as radioactive isotopes) that decompose at a rate proportional to the amount \(Q\) present at any instant \(t\).

Exponential Decay Model

\[ \dfrac{dQ}{dt} = -kQ \implies Q(t) = Q_0 e^{-kt} \]

Example. Radium decomposes at a rate proportional to the amount present. If half of the original amount disappears after \(1000\text{ years}\), what percentage is lost in \(100\text{ years}\)?

Solution. Apply \(Q(t) = Q_0 e^{-kt}\). At \(t = 1000\), \(Q = 0.5Q_0\):

\[ 0.5Q_0 = Q_0 e^{-1000k} \implies -1000k = \ln 0.5 \implies k = 0.000693 \]

Calculate the amount remaining at \(t = 100\text{ years}\):

\[ Q(100) = Q_0 e^{-0.000693(100)} = Q_0 e^{-0.0693} \approx 0.933Q_0 \]

Calculate percentage lost:

\[ \text{Percentage Lost} = \dfrac{Q_0 - 0.933Q_0}{Q_0} \times 100\% = 6.7\% \]

Answer: \(6.7\%\)

Applications of DE: Cooling and Heating Problems

Newton’s Law of Cooling states that the rate of change of temperature \(T\) of a body is proportional to the difference between its temperature and the surrounding ambient temperature \(T_s\).

Newton’s Law of Cooling Formula

\[ \dfrac{dT}{dt} = -k(T - T_s) \implies T(t) = T_s + (T_0 - T_s)e^{-kt} \]

Example. Air temperature is \(30^\circ\text{C}\) and a body cools from \(100^\circ\text{C}\) to \(70^\circ\text{C}\) in \(15\text{ minutes}\). How long will it take to cool from \(100^\circ\text{C}\) to \(50^\circ\text{C}\)?

Solution. Here \(T_s = 30^\circ\text{C}\) and \(T_0 = 100^\circ\text{C}\). The model is \(T(t) - 30 = 70e^{-kt}\). At \(t = 15\), \(T = 70\):

\[ 70 - 30 = 70e^{-15k} \implies \dfrac{40}{70} = e^{-15k} \implies k = \dfrac{-\ln(4/7)}{15} \approx 0.0373 \]

Find \(t\) when \(T = 50^\circ\text{C}\):

\[ 50 - 30 = 70e^{-0.0373 t} \implies \dfrac{20}{70} = e^{-0.0373 t} \implies t = \dfrac{-\ln(2/7)}{0.0373} \approx 33.59\text{ minutes} \]

Answer: \(33.59\text{ minutes}\)

Applications of DE: Flow Problems

Flow or mixture problems calculate the instantaneous amount of salt or solute \(Q(t)\) in a tank using the principle that the net rate of change equals the rate of gain minus the rate of loss.

Flow / Mixture Rate Differential Equation

\[ \dfrac{dQ}{dt} = \text{Rate of Gain} - \text{Rate of Loss} = R_{\text{in}} C_{\text{in}} - R_{\text{out}} \left(\dfrac{Q}{V(t)}\right) \]

Example. A tank contains \(100\text{ liters}\) of brine holding \(50\text{ kg}\) of salt. Pure water enters at \(3\text{ L/min}\) and brine flows out at \(2\text{ L/min}\). How much salt remains after \(1\text{ hour}\) (\(60\text{ minutes}\))?

Solution. Volume at time \(t\) is \(V(t) = 100 + (3 - 2)t = 100 + t\). Rate of gain is \(3 \times 0 = 0\). Rate of loss is \(2 \times \dfrac{Q}{100 + t}\):

\[ \dfrac{dQ}{dt} = -\dfrac{2Q}{100 + t} \implies \dfrac{dQ}{Q} = -\dfrac{2dt}{100 + t} \]

Integrate both sides:

\[ \ln Q = -2\ln(100 + t) + C \implies Q(t) = \dfrac{K}{(100 + t)^2} \]

At \(t = 0, Q = 50 \implies 50 = \dfrac{K}{100^2} \implies K = 500,000\). Substitute \(t = 60\text{ min}\):

\[ Q(60) = \dfrac{500,000}{(160)^2} = \dfrac{500,000}{25,600} \approx 19.53\text{ kg} \]

Answer: \(19.53\text{ kg}\)

Applications of DE: Continuous Compound Interest

Continuous compound interest assumes the account balance \(P(t)\) grows at a rate proportional to the current balance with continuous nominal annual interest rate \(r\).

Continuous Compound Interest Model

\[ \dfrac{dP}{dt} = rP \implies P(t) = P_0 e^{rt} \]

Example. An initial principal of $P_0 = \$1,000$ is invested at a nominal continuous interest rate of \(6\%\) per year. Determine the balance after \(10\text{ years}\).

Solution. Set up the differential equation \(\dfrac{dP}{dt} = 0.06P\). Solving yields \(P(t) = 1000e^{0.06t}\). At \(t = 10\text{ years}\):

\[ P(10) = 1000 e^{0.06(10)} = 1000 e^{0.6} \approx 1000(1.82212) = \$1,822.12 \]

Answer: $\$1,822.12$

Applications of DE: Motion Problems

Motion problems express velocity \(v\) and acceleration \(a\) as derivatives of displacement \(s\) with respect to time \(t\).

Kinematic Differential Equations

\[ v = \dfrac{ds}{dt}, \qquad\qquad a = \dfrac{dv}{dt} = v\dfrac{dv}{ds} \]

Example. An object falls from rest in a resisting medium. Its velocity before reaching the ground is given by \(\dfrac{dV}{dt} + \dfrac{V}{10} = 32\text{ ft/sec}^2\). Find the velocity \(1\text{ second}\) after it falls.

Solution. Rearrange and separate variables:

\[ 10\dfrac{dV}{dt} + V = 320 \implies \dfrac{dV}{320 - V} = \dfrac{dt}{10} \]

Integrate both sides:

\[ -\ln(320 - V) = \dfrac{t}{10} + c \implies 320 - V = K e^{-t/10} \]

At \(t = 0, V = 0 \implies K = 320 \implies V(t) = 320(1 - e^{-t/10})\). At \(t = 1\text{ sec}\):

\[ V(1) = 320(1 - e^{-0.1}) = 320(1 - 0.904837) \approx 30.45\text{ ft/sec} \]

Answer: \(30.45\text{ ft/sec}\)

Applications of DE: Newton’s Second Law

Newton’s Second Law of Motion relates net force \(F\), mass \(m\), and acceleration \(a = \dfrac{dv}{dt}\).

Newton’s Second Law Differential Equation

\[ F = m\dfrac{dv}{dt} = m\dfrac{d^2s}{dt^2} \]

Example. A mass \(m = 2\text{ kg}\) is acted upon by a constant force \(F = 10\text{ N}\). If the body starts from rest (\(v = 0\) at \(t = 0\)), find its velocity at \(t = 3\text{ seconds}\).

Solution. Set up Newton’s second law differential equation:

\[ 10 = 2\dfrac{dv}{dt} \implies \dfrac{dv}{dt} = 5 \]

Integrate both sides with respect to time \(t\):

\[ v(t) = 5t + C \]

Since \(v(0) = 0\), \(C = 0 \implies v(t) = 5t\). At \(t = 3\text{ seconds}\):

\[ v(3) = 5(3) = 15\text{ m/s} \]

Answer: \(15\text{ m/s}\)

Applications of DE: Orthogonal Trajectories

An orthogonal trajectory is a curve that intersects every member of a given family of curves at right angles (\(90^\circ\)). Their slopes are negative reciprocals of each other.

Orthogonal Trajectory Slope Condition

\[ \left(\dfrac{dy}{dx}\right)_{\text{ortho}} = -\dfrac{1}{\left(\dfrac{dy}{dx}\right)_{\text{given}}} \]

Example. Find the equation of the family of orthogonal trajectories of the system of parabolas \(y^2 = 2x + C\).

Solution. Differentiate \(y^2 = 2x + C\) to find the slope of the given family:

\[ 2y\dfrac{dy}{dx} = 2 \implies \left(\dfrac{dy}{dx}\right)_{\text{given}} = \dfrac{1}{y} \]

The slope of the orthogonal trajectories is the negative reciprocal:

\[ \left(\dfrac{dy}{dx}\right)_{\text{ortho}} = -y \implies \dfrac{dy}{y} = -dx \]

Integrate both sides:

\[ \ln y = -x + c \implies y = C e^{-x} \]

Answer: \(y = C e^{-x}\)

Formula Summary

Formula Name Formula
Variable Separable Form \(P(x)dx+Q(y)dy=0 \implies \int P(x)dx+\int Q(y)dy=C\)
Homogeneous Substitution \(y=vx, \quad dy=v\,dx+x\,dv\)
Linear First-Order DE \(\dfrac{dy}{dx}+P(x)y=Q(x)\)
First-Order Linear Solution \(y \cdot e^{\int P(x)dx}=\int Q(x)e^{\int P(x)dx}dx+C\)
Second-Order Auxiliary Equation \(am^2+bm+c=0\)
Real & Distinct Roots Solution \(y=C_1 e^{m_1 x}+C_2 e^{m_2 x}\)
Real & Equal Roots Solution \(y=(C_1 x+C_2)e^{mx}\)
Complex Conjugate Roots Solution \(y=e^{\alpha x}(C_1 \cos \beta x+C_2 \sin \beta x)\)
Exactness Condition \(\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}\)
Bernoulli DE Transformation \(v=y^{1-n} \implies \dfrac{dv}{dx}+(1-n)P(x)v=(1-n)Q(x)\)
Elimination of Constants \(\text{Differentiations}=\text{Number of Constants}\)
Exponential Growth & Decay \(P(t)=P_0 e^{kt} \quad \text{or} \quad Q(t)=Q_0 e^{-kt}\)
Newton’s Law of Cooling \(T(t)=T_s+(T_0-T_s)e^{-kt}\)
Mixture Flow Rate Equation \(\dfrac{dQ}{dt}=R_{\text{in}}C_{\text{in}}-R_{\text{out}}\left(\dfrac{Q}{V(t)}\right)\)
Continuous Compound Interest \(P(t)=P_0 e^{rt}\)
Kinematic Differential Equations \(v=\dfrac{ds}{dt}, \quad a=\dfrac{dv}{dt}=v\dfrac{dv}{ds}\)
Newton’s Second Law of Motion \(F=m\dfrac{dv}{dt}=m\dfrac{d^2s}{dt^2}\)
Orthogonal Trajectory Condition \(\left(\dfrac{dy}{dx}\right)_{\text{ortho}}=-\dfrac{1}{\left(\dfrac{dy}{dx}\right)_{\text{given}}}\)