izak

applied differential equations review

Welcome to my review for the class that I always forget the most about while I am tutoring.

2.1: Linear First Order Equations

Solve \(y' + 2y = e^{-x}\). The integrating factor \(\mu(x) = e^{2x}\) yields the solution \(y(x) = \frac{1}{3}e^{-x} + Ce^{-2x}\).

2.2: Separable Equations

Solve \(y' = xy^2\). Separating variables and integrating gives \(y(x) = -\frac{2}{x^2 + C}\).

2.3: Existence and Uniqueness of Nonlinear Equations

If a function \(f(x, y)\) is continuous within an open rectangle in the \(xy\)-plane containing a point \((x_0, y_0)\), then the initial value problem \(y' = f(x, y)\), \(y(x_0) = y_0\) has at least one solution (existence). If the partial derivative \(\frac{\partial f}{\partial y}\) is also continuous within that rectangle, then the solution is unique.

2.4: Transformation into Separable Equations

To solve the differential equation \(y' = \frac{x + y}{x - y}\), we use the substitution \(v = \frac{y}{x}\), which implies \(y = vx\) and \(y' = v + xv'\). Substituting into the equation gives \(v + xv' = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}.\) Rearranging, we get \(xv' = \frac{1 + v}{1 - v} - v = \frac{1 + v^2}{1 - v}.\) This is now a separable equation in \(v\) and \(x\). Separating variables and integrating yields \(\int \frac{1 - v}{1 + v^2} \, dv = \int \frac{1}{x} \, dx.\) Solving these integrals and substituting back \(v = \frac{y}{x}\) gives the solution to the original equation.

2.5: Exact Equations

Solve \((2xy + y^2)dx + (x^2 + 2xy)dy = 0\). The potential function \(\Psi(x, y) = x^2y + xy^2\) yields the solution \(x^2y + xy^2 = C\).

2.6: Integrating Factors

Solve \(y dx - x dy = 0\). The integrating factor \(\mu(x, y) = \frac{1}{x^2}\) makes it exact, yielding \(\frac{y}{x} = C\).

4.1: Growth and Decay

A population grows exponentially: \(P(t) = P_0 e^{kt}\), where \(P_0\) is the initial population and \(k\) is the growth rate.

4.2: Cooling and Mixing

Newton’s Law of Cooling: \(T(t) = T_a + (T_0 - T_a)e^{-kt}\), where \(T_a\) is ambient temperature and \(T_0\) is initial temperature.

4.3: Elementary Mechanics

A falling object with air resistance: \(m\frac{dv}{dt} = mg - kv\). The solution is \(v(t) = \frac{mg}{k}(1 - e^{-kt/m})\).

4.4: Autonomous Second Order Equations

Solve \(y'' + y = 0\). The general solution is \(y(x) = C_1 \cos x + C_2 \sin x\).

5.2: Constant Coefficient Homogeneous Equations

A constant coefficient homogeneous equation has the form \(y'' + ay' + by = 0\), where \(a\) and \(b\) are constants. The solution is found by solving the characteristic equation \(r^2 + ar + b = 0\). For \(y'' + 4y' + 4y = 0\), the characteristic equation \(r^2 + 4r + 4 = 0\) yields a repeated root \(r = -2\). This results in the general solution \(y(x) = (C_1 + C_2 x)e^{-2x}\), where \(C_1\) and \(C_2\) are constants determined by initial conditions.

5.3: Nonhomogeneous Linear Equations

A nonhomogeneous linear equation has the form \(y'' + ay' + by = g(x)\), where \(g(x)\) is a nonzero function. The general solution is the sum of the complementary solution (solution to the homogeneous equation) and a particular solution to the nonhomogeneous equation. For \(y'' + y = \sin x\), the complementary solution is \(y_c(x) = C_1 \cos x + C_2 \sin x\), and a particular solution is \(y_p(x) = -\frac{1}{2}x \cos x\). Thus, the general solution is \(y(x) = C_1 \cos x + C_2 \sin x - \frac{1}{2}x \cos x\).

5.4: Method of Undetermined Coefficients

The method of undetermined coefficients is a technique for finding a particular solution \(y_p(x)\) to nonhomogeneous linear differential equations of the form \(y'' + ay' + by = g(x)\), where \(g(x)\) is a polynomial, exponential, sine, cosine, or a combination of these. The idea is to guess a form for \(y_p(x)\) based on \(g(x)\), with undetermined coefficients, and then substitute it into the equation to solve for those coefficients. For \(y'' + y = x^2\), since \(g(x) = x^2\) is a polynomial, we guess \(y_p(x) = Ax^2 + Bx + C\). Substituting \(y_p\), \(y_p'\), and \(y_p''\) into the equation and equating coefficients, we find \(A = 1\), \(B = 0\), and \(C = -2\), giving \(y_p(x) = x^2 - 2\). The general solution combines this particular solution with the complementary solution \(y_c(x) = C_1 \cos x + C_2 \sin x\), resulting in \(y(x) = C_1 \cos x + C_2 \sin x + x^2 - 2\). This method works efficiently for specific types of \(g(x)\).

5.5: Method of Undetermined Coefficients II

The method of undetermined coefficients extends to cases where \(g(x)\) involves products of exponentials, sines, and cosines. For \(y'' + y = e^x \cos x\), we guess a particular solution of the form \(y_p(x) = e^x (A \cos x + B \sin x)\), where \(A\) and \(B\) are undetermined coefficients. Substituting \(y_p\), \(y_p'\), and \(y_p''\) into the equation and equating coefficients of \(e^x \cos x\) and \(e^x \sin x\), we solve for \(A\) and \(B\). This yields \(A = \frac{1}{5}\) and \(B = \frac{2}{5}\), giving the particular solution \(y_p(x) = \frac{1}{5}e^x (\cos x + 2 \sin x)\). The general solution combines this with the complementary solution \(y_c(x) = C_1 \cos x + C_2 \sin x\), resulting in \(y(x) = C_1 \cos x + C_2 \sin x + \frac{1}{5}e^x (\cos x + 2 \sin x)\). This method is effective for handling more complex forms of \(g(x)\).

5.6: Reduction of Order

Reduction of order is a method used to find a second linearly independent solution \(y_2(x)\) to a second-order linear homogeneous differential equation when one solution \(y_1(x)\) is already known. For \(y'' - 2y' + y = 0\), given \(y_1 = e^x\), we assume \(y_2 = v(x)e^x\), where \(v(x)\) is an unknown function. Substituting \(y_2\), \(y_2'\), and \(y_2''\) into the equation and simplifying, we find \(v''(x) = 0\), which implies \(v(x) = C_1 + C_2 x\). Choosing \(C_1 = 0\) and \(C_2 = 1\) gives the second solution \(y_2(x) = xe^x\). We choose \(C_1 = 0\) and \(C_2 = 1\) to ensure \(y_2(x) = v(x)e^x\) is linearly independent from the known solution \(y_1(x) = e^x\). Setting \(C_1 = 0\) avoids redundancy, as including it would introduce a term proportional to \(y_1(x)\). Choosing \(C_2 = 1\) simplifies the solution, making \(y_2(x) = xe^x\) the simplest linearly independent solution. Together, \(y_1(x) = e^x\) and \(y_2(x) = xe^x\) form a basis for the general solution \(y(x) = C_1 e^x + C_2 xe^x\). This method reduces the problem to solving a first-order equation for \(v'(x)\).

5.7: Variation of Parameters

Variation of parameters is a method for finding a particular solution to a nonhomogeneous linear differential equation \(y'' + p(x)y' + q(x)y = g(x)\). For \(y'' + y = \tan x\), the complementary solution is \(y_c(x) = C_1 \cos x + C_2 \sin x\). We seek a particular solution of the form \(y_p(x) = u_1(x) \cos x + u_2(x) \sin x\), where \(u_1(x)\) and \(u_2(x)\) are functions to be determined. For the equation \(y'' + y = \tan x\), the complementary solution is \(y_c(x) = C_1 \cos x + C_2 \sin x\). For first constraint, to simplify the problem, we impose the condition:$ u_1’(x) x + u_2’(x) x = 0.$ This ensures that the derivative of \(y_p(x)\) does not include terms involving \(u_1'(x)\) and \(u_2'(x)\) beyond what is necessary. For second constraint, substitute \(y_p(x)\), \(y_p'(x)\), and \(y_p''(x)\) into the original equation \(y'' + y = \tan x\). After simplification, this yields:$ -u_1’(x) x + u_2’(x) x = x.$ Solving the system \(u_1'(x) \cos x + u_2'(x) \sin x = 0\) and \(-u_1'(x) \sin x + u_2'(x) \cos x = \tan x\), we find \(u_1'(x) = -\sin x \tan x\) and \(u_2'(x) = \sin x\). Integrating these gives \(u_1(x) = \ln|\sec x + \tan x| - \sin x\) and \(u_2(x) = -\cos x\). Thus, the particular solution is \(y_p(x) = -\cos x \ln|\sec x + \tan x|\), and the general solution is \(y(x) = C_1 \cos x + C_2 \sin x - \cos x \ln|\sec x + \tan x|\). This method is more general than undetermined coefficients and works for any continuous \(g(x)\).

7.4: Series Solutions Near an Ordinary Point II

Consider this example from William Trench “Elementary Differential Equations & Boundary Value Problems” section \(7.4\). Find the coefficients \(a_0, \dots, a_7\) in the series solution \(y = \sum_{n=0}^\infty a_n x^n\) of the initial value problem \((1 + x + 2x^2)y'' + (1 + 7x)y' + 2y = 0, \quad y(0) = -1, \quad y'(0) = -2. \quad (*)\) Here, the differential operator is \(Ly = (1 + x + 2x^2)y'' + (1 + 7x)y' + 2y\). The zeros of \(P_0(x) = 1 + x + 2x^2\), given by \(\frac{-1 \pm i\sqrt{7}}{4}\), have absolute value \(\frac{1}{\sqrt{2}}\). Now if \(P_0(x)\) has no zeros in the interval \(|x| < R\), then the series solution converges on \(|x| < R\). So the series solution converges on \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\). Substituting \(y = \sum_{n=0}^\infty a_n x^n\), \(y' = \sum_{n=1}^\infty n a_n x^{n-1}\), and \(y'' = \sum_{n=2}^\infty n(n-1) a_n x^{n-2}\) into \(Ly\), we obtain \(Ly = \sum_{n=2}^\infty n(n-1) a_n x^{n-2} + \sum_{n=2}^\infty n(n-1) a_n x^{n-1} + 2\sum_{n=2}^\infty n(n-1) a_n x^n + \sum_{n=1}^\infty n a_n x^{n-1} + 7\sum_{n=1}^\infty n a_n x^n + 2\sum_{n=0}^\infty a_n x^n.\) Shifting indices so the general term in each series is a constant multiple of \(x^n\) yields \(Ly = \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n + \sum_{n=0}^\infty (n+1)n a_{n+1} x^n + 2\sum_{n=0}^\infty n(n-1) a_n x^n + \) \(\sum_{n=0}^\infty (n+1) a_{n+1} x^n + 7\sum_{n=0}^\infty n a_n x^n + 2\sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty b_n x^n,\) where \(b_n = (n+2)(n+1) a_{n+2} + (n+1)^2 a_{n+1} + (n+2)(2n+1) a_n.\) Thus, \(y = \sum_{n=0}^\infty a_n x^n\) is a solution of \(Ly = 0\) if and only if \(a_{n+2} = -\frac{n+1}{n+2} a_{n+1} - \frac{2n+1}{n+1} a_n, \quad n \geq 0. \quad (**)\) From the initial conditions, \(a_0 = y(0) = -1\) and \(a_1 = y'(0) = -2\). Setting \(n = 0\) in Equation \((**)\) yields \(a_2 = -\frac{1}{2} a_1 - a_0 = -\frac{1}{2}(-2) - (-1) = 2.\) Setting \(n = 1\) in Equation \((**)\) yields \(a_3 = -\frac{2}{3} a_2 - \frac{3}{2} a_1 = -\frac{2}{3}(2) - \frac{3}{2}(-2) = \frac{5}{3}.\) The coefficients \(a_4, a_5, a_6, a_7\) can be computed similarly from Equation \((**)\), yielding the series solution \(y = -1 - 2x + 2x^2 + \frac{5}{3}x^3 - \frac{55}{12}x^4 + \frac{3}{4}x^5 + \frac{61}{8}x^6 - \frac{443}{56}x^7 + \cdots.\) The Taylor polynomials \(T_N(x) = \sum_{n=0}^N a_n x^n\) converge to the solution of Equation \((*)\) on \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\).

8.1: Introduction to the Laplace Transform

The Laplace transform of a function \(f(t)\), defined for \(t \geq 0\), is given by \(\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) \, dt,\) where \(s\) is a complex variable. It converts functions of time \(t\) into functions of the complex variable \(s\). For example, the Laplace transform of \(e^{at}\) is \(\mathcal{L}\{e^{at}\} = \frac{1}{s - a}, \quad \text{for } s > a.\) In practice, Laplace transforms of common functions are often identified using precalculated tables, which provide a quick reference for solving problems without repeatedly computing the integral.

8.2: The Inverse Laplace Transform

Find \(\mathcal{L}^{-1}\left\{\frac{1}{s^2 + 4}\right\} = \frac{1}{2} \sin 2t\).

8.3: Solution of Initial Value Problems

Solve \(y'' + y = 0\), \(y(0) = 1\), \(y'(0) = 0\). The Laplace transform yields \(y(t) = \cos t\).

8.4 The Unit Step Function

The unit step function \(u(t - a)\) is defined as 0 for \(t < a\) and 1 for \(t \geq a\). For the equation \(y'' + y = u(t - \pi)\) with \(y(0) = 0\) and \(y'(0) = 0\), we apply the Laplace transform. The Laplace transform of \(u(t - \pi)\) is \(\frac{e^{-\pi s}}{s}\). Taking the Laplace transform of the equation gives \(s^2 Y(s) + Y(s) = \frac{e^{-\pi s}}{s}.\) Solving for \(Y(s)\), we get \(Y(s) = \frac{e^{-\pi s}}{s(s^2 + 1)}.\) Using partial fractions and the inverse Laplace transform, the solution is \(y(t) = u(t - \pi)(1 - \cos(t - \pi)).\)

8.5 Piecewise Continuous Forcing Functions

For \(y'' + y = f(t)\), where \(f(t) = 1\) for \(0 \leq t < \pi\) and \(0\) otherwise, we express \(f(t)\) using the unit step function as \(f(t) = u(t) - u(t - \pi)\). The Laplace transform of \(f(t)\) is \(F(s) = \frac{1}{s} - \frac{e^{-\pi s}}{s}.\) Taking the Laplace transform of the equation \(y'' + y = f(t)\) gives \(s^2 Y(s) + Y(s) = \frac{1}{s} - \frac{e^{-\pi s}}{s}.\) Solving for \(Y(s)\), we obtain \(Y(s) = \frac{1}{s(s^2 + 1)} - \frac{e^{-\pi s}}{s(s^2 + 1)}.\) Using partial fractions and the inverse Laplace transform, the solution is \(y(t) = (1 - \cos t) - u(t - \pi)(1 - \cos(t - \pi)).\)

8.6 Convolution

For \(y'' + y = \sin t\) with \(y(0) = 0\) and \(y'(0) = 0\), we use the convolution integral. The impulse response of the system is \(h(t) = \sin t\). The solution is given by the convolution of \(\sin t\) with the input \(\sin t\) \(y(t) = \int_0^t \sin(t - \tau) \sin \tau \, d\tau.\) Using the trigonometric identity \(\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]\), the integral becomes \(y(t) = \frac{1}{2} \int_0^t [\cos(t - 2\tau) - \cos t] \, d\tau.\) Evaluating the integral, we obtain \(y(t) = \frac{1}{2}(\sin t - t \cos t).\)

8.7 Impulse Functions

For \(y'' + y = \delta(t - \pi)\) with \(y(0) = 0\) and \(y'(0) = 0\), the Laplace transform of \(\delta(t - \pi)\) is \(e^{-\pi s}\). Taking the Laplace transform of the equation gives \(s^2 Y(s) + Y(s) = e^{-\pi s}.\) Solving for \(Y(s)\), we get \(Y(s) = \frac{e^{-\pi s}}{s^2 + 1}.\) The inverse Laplace transform yields \(y(t) = u(t - \pi) \sin(t - \pi).\)