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# Differential Equations 1st Midterm Study Guide AS.110.302

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This 5 page Study Guide was uploaded by Laura Chicos on Friday October 7, 2016. The Study Guide belongs to AS.110.302 at Johns Hopkins University taught by Dr. Yannick Sire in Fall 2016. Since its upload, it has received 10 views.

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Date Created: 10/07/16

Differential Equations Exam I Study Guide st DE Study Guide for 1 Midterm: Chapters 2.1 – 3.6 2.1 – Linear Equations; Method of Integrating Factors Standard form of first order linear equation: dy/dt + p(t)y = g(t) Method of integrating factors o put given equation into standard form if it is not o multiply whole equation by a function (t) o generally, (t) = e^(t/2) o more specifically, (t) = e^p(t)dt o now, left side should be the derivative of some product o integrate both sides o to find general solution, solve equation for y o if there is an initial value, plug in point to find constant, c Note: may have to use integration by parts if there is a term like te^t Integration by parts: udv = uv - vdu 2.2 – Separable Equations We wave our hand &… Separate the variables and integrate both sides E.g. dy/dt = t/(y+1) (y+1)dy = tdt o don’t forget the integration constant Implicit solution: have the original equation with different degrees of y Explicit solution: solve for y using quadratic formula (not always easy) to find interval of validity: o find values of independent variable (usually t), where the function, y, is continuous o when given an initial condition, y(t0) = x, see upon which interval of continuity t0 lies, and that is the interval upon which the equation has a unique solution o don’t forget u-sub as a tool 2.4 – Existence and Uniqueness Theorem Theorem 2.4.1: If the functions p and g are continuous on an open interval I: alpha<t<beta, containing the point t=t0, then there exists a unique function y=phi(t) that satisfies the differential equation dy/dt + p(t)y = g(t) for each t in I, and that also satisfies the initial condition y(t0) = y0, where y0 is an arbitrary prescribed initial value. Theorem 2.4.2: Let the functions f and df/dy be continuous in some rectangle a<t<b, c<y<d containing the point (t0,y0). Then, in some interval t0-h<t<t0+h contained in a<t<b, there is a unique solution y=phi(t) of the initial value problem y’=f(t,y), y(t0) = y0 2.5 – Autonomous Equations Autonomous equation – equation in which the independent variable does not appear explicitly Autonomous equations have the form dy/dt = f(y) can sketch vector field in ty-plane to find equilibrium points: set function equal to 0 and solve for y Asymptotically stable solution: attractive above and below the critical point or in other words, solution approaches the equilibrium solution asymptotically as tinfinity Unstable Equilibrium solution: repulsive above and below the critical point, or the solution diverges from the equilibrium point Semi-stable Equilibrium Solution: solutions lying on one side of equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it 2.6 – Exact Equations if a differential equation is neither linear nor separable, then it is most likely exact M(x,y) + N(x,y)y’ = 0 d/dx(x,y) = M(x,y), d/dy(x,y) = N(x,y) such that (x,y) = c defines y = phi(x) implicitly as a differentiable function of x To solve an exact differential equation: o calculate the partials My and Nx o if the partials are equal, then the equation is exact o x(x,y) = M(x,y) and y(x,y) = N(x,y) o integrate x(x,y) = M(x,y) to get (x,y) o Note: when taking the integral, make sure to include an arbitrary function h(y) rather than an arbitrary constant o Then take the partial of (x,y) with respect to y and set it equal to N(x,y) o In doing the latter step, you can solve for h’(y), which you can then integrate to find h(y) o substitue for h(y) in (x,y) o Solution is given implicitly by setting (x,y) = c. Note: it is sometimes possible to convert a differential equation that is not exact into an exact equation by multiplying the equation by a suitable integrating factor (x) can be found by solving the following equation which is both linear and separable: Then, multiply original equation by (x) and check that the equation is exact, which now it should be. Continue to solve in the same way 2.7 – Euler Method Equation for the line tangent to the solution curve at (t0,y0), namely, y1 = y0 + f(t0,y0)(t1-t0) Repeat the process again to find the next y, by replacing y0 with y1 and so on If we assume that there is a uniform step size h between the points t0, t1, t2… then we obtain Euler’s formula in the form: y n+1 = y n f hn n=0,1,2…. 3.1 – Homogenous Equations with Constant Coefficients Homogeneous equation means the equation is equal to 0. First, find characteristic equation Find the roots If there are two distinct and real roots, 1 and 2, the general solution is y=c1e^1t + c2e^2t 3.2 – General Theory on Linear Homogeneous Equations linear equations: o y’’ + p(t)y’ + q(t)y = g(t) o P(t)y’’ + Q(t)y’ + R(t)y = G(t) homogeneous equation: terms g(t) or G(t) are zero for all t. Theorem 3.2.1: (Existence and Uniqueness Theorem): o Consider the standard initial value problem y’’ + p(t)y’ + q(t)y = g(t), y(t0) = y0, y’(t0) = y’0, where p,q and g are continuous & at least twice differentiable on an open interval I that contains the point t0. Then there is exactly one solution y= phi(t) of this problem, and the solution exists throughout the interval I. Theorem 3.2.2: ( Principle of Superposition): o If y1 and y2 are two solutions of the differential equation L[y] = y’’ + p(t)y’ + q(t)y = 0, then the linear combination c1y1 + c2y2 is also a solution for any values of the constants c1 and c2. The Wronskian: y1y’2 – y’1y2 o Two solutions of a homogenous equation are said to form a fundamental set of solutions if their Wronskian is nonzero If and only in this case, their linear combination includes all solutions of the homogenous equation, and we then call this linear combination the general solution. 3.3 – Complex, Conjugate Roots Characteristic equation has complex, conjugate roots Roots are denoted by: r1 = + i, r2 = - i Calculate the Wronskian to make sure that y1 and y2 form a fundamental set of solutions General solution: y= c1e^cost + c2e^sint 3.4 – Repeated Roots; Reduction of Order Characteristic equation has repeated, real roots One solution of the differential equation is y1(t) = e^(r1t) To find the general solution, we need a second solution that is not a constant multiple of y1 Reduction of Order: So, replace c by the function v(t) and try to determine v(t) so that the product v(t)y1(t) is also a solution of the differential equation To do this, substitute y = v(t)y1(t) into original differential equation (i.e. take first and second derivatives and plug it in) and use the resulting equation to find v(t) Second solution will look like y2(t) = t*exp(r1t) Verify that these two solutions form a fundamental set by calculating their Wronskian 3.5 – Non-homogeneous Equations; Method of Undetermined Coefficients Theorem 3.5.2: o Find the general solution c1y1(t) + c2y2(t) of the corresponding homogeneous equation. This solution is called the complementary solution and can be denoted as yc(t) o Find some single solution, i.e. a particular solution, Y(t) of the non-homogeneous equation. o Form the sum of the complementary and particular solutions The Particular Solutions of the right-hand side of the non- homogeneous equation gi(t) Yi(t) polynomial t^s(A0t^n +A1t^(n-1) +…_An) Ae^at t^s(A0t^n….An)e^at A(e^at)sinBt or t^s*A*e^(at)cosBt + B(e^at)cosBt B*e^(at)sinBt Note: s is the smallest nonnegative integer that will ensure that no term in the particular solution, Yi(t) is a solution of the corresponding homogeneous solution. 3.6 – Variation of Parameters Basic Idea: replace the constants c1 and c2 with the functions u1(t) and u2(t) Particular Solution, Y(t) = y1(t)u1(t) +y2(t)u2(t) General solution: y = yc+yp (homogeneous solution plus particular solution)

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