Calculus Volume 3

# Key Concepts

Calculus Volume 3Key Concepts

### 7.1Second-Order Linear Equations

• Second-order differential equations can be classified as linear or nonlinear, homogeneous or nonhomogeneous.
• To find a general solution for a homogeneous second-order differential equation, we must find two linearly independent solutions. If $y1(x)y1(x)$ and $y2(x)y2(x)$ are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by
$y(x)=c1y1(x)+c2y2(x).y(x)=c1y1(x)+c2y2(x).$
• To solve homogeneous second-order differential equations with constant coefficients, find the roots of the characteristic equation. The form of the general solution varies depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.
• Initial conditions or boundary conditions can then be used to find the specific solution to a differential equation that satisfies those conditions, except when there is no solution or infinitely many solutions.

### 7.2Nonhomogeneous Linear Equations

• To solve a nonhomogeneous linear second-order differential equation, first find the general solution to the complementary equation, then find a particular solution to the nonhomogeneous equation.
• Let $yp(x)yp(x)$ be any particular solution to the nonhomogeneous linear differential equation
$a2(x)y″+a1(x)y′+a0(x)y=r(x),a2(x)y″+a1(x)y′+a0(x)y=r(x),$

and let $c1y1(x)+c2y2(x)c1y1(x)+c2y2(x)$ denote the general solution to the complementary equation. Then, the general solution to the nonhomogeneous equation is given by
$y(x)=c1y1(x)+c2y2(x)+yp(x).y(x)=c1y1(x)+c2y2(x)+yp(x).$
• When $r(x)r(x)$ is a combination of polynomials, exponential functions, sines, and cosines, use the method of undetermined coefficients to find the particular solution. To use this method, assume a solution in the same form as $r(x),r(x),$ multiplying by x as necessary until the assumed solution is linearly independent of the general solution to the complementary equation. Then, substitute the assumed solution into the differential equation to find values for the coefficients.
• When $r(x)r(x)$ is not a combination of polynomials, exponential functions, or sines and cosines, use the method of variation of parameters to find the particular solution. This method involves using Cramer’s rule or another suitable technique to find functions $u′(x)u′(x)$ and $v′(x)v′(x)$ satisfying
$u′y1+v′y2=0u′y1′+v′y2′=r(x).u′y1+v′y2=0u′y1′+v′y2′=r(x).$

Then, $yp(x)=u(x)y1(x)+v(x)y2(x)yp(x)=u(x)y1(x)+v(x)y2(x)$ is a particular solution to the differential equation.

### 7.3Applications

• Second-order constant-coefficient differential equations can be used to model spring-mass systems.
• An examination of the forces on a spring-mass system results in a differential equation of the form
$mx″+bx′+kx=f(t),mx″+bx′+kx=f(t),$

where $mm$ represents the mass, $bb$ is the coefficient of the damping force, $kk$ is the spring constant, and $f(t)f(t)$ represents any net external forces on the system.
• If $b=0,b=0,$ there is no damping force acting on the system, and simple harmonic motion results. If $b≠0,b≠0,$ the behavior of the system depends on whether $b2−4mk>0,b2−4mk>0,$ $b2−4mk=0,b2−4mk=0,$ or $b2−4mk<0.b2−4mk<0.$
• If $b2−4mk>0,b2−4mk>0,$ the system is overdamped and does not exhibit oscillatory behavior.
• If $b2−4mk=0,b2−4mk=0,$ the system is critically damped. It does not exhibit oscillatory behavior, but any slight reduction in the damping would result in oscillatory behavior.
• If $b2−4mk<0,b2−4mk<0,$ the system is underdamped. It exhibits oscillatory behavior, but the amplitude of the oscillations decreases over time.
• If $f(t)≠0,f(t)≠0,$ the solution to the differential equation is the sum of a transient solution and a steady-state solution. The steady-state solution governs the long-term behavior of the system.
• The charge on the capacitor in an RLC series circuit can also be modeled with a second-order constant-coefficient differential equation of the form
$Ld2qdt2+Rdqdt+1Cq=E(t),Ld2qdt2+Rdqdt+1Cq=E(t),$

where L is the inductance, R is the resistance, C is the capacitance, and $E(t)E(t)$ is the voltage source.

### 7.4Series Solutions of Differential Equations

• Power series representations of functions can sometimes be used to find solutions to differential equations.
• Differentiate the power series term by term and substitute into the differential equation to find relationships between the power series coefficients.