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DiffEq Notes2

# DiffEq Notes2 - Day 8 In Day 8 we began Chapters 4(second...

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Day 8: In Day 8 we began Chapters 4 (second order linear) and 6 (linear higher than second order ). Since differential equations in both chapters are handled exactly the same way, both will be studied at the same time and homework from both chapters can be done at the same time. There are precisely two kinds of equations that we will learn to solve, constant coefficient (the coefficients are constants - have no x's) and Cauchy-Euler (the coefficients are a constant times a power of x, where the power exactly matches the order of the derivative of y for which it is a coefficient. Here are examples of each of the two types, taken from p. 15 in the HW. Both are homogeneous, meaning that the right hand side is zero. We will begin with homogeneous equations, since the solutions to linear equations that are not homogeneous are built from the homogeneous solutions. Nonhomogeneous solutions will be covered next week. Type: Linear homogeneous (rhs = 0) constant coefficient equation. > restart: > de:=diff(y(x),x\$3)+3*diff(y(x),x\$2)+2*diff(y(x),x)=0; / 3 \ / 2 \ |d | |d | /d \ de := |--- y(x)| + 3 |--- y(x)| + 2 |-- y(x)| = 0 | 3 | | 2 | \dx / \dx / \dx / The coefficients are 1, 3, and 2, all of which are constants. For constant coefficient equations, the form of an explicit solution is always y(x) = exp(m*x). > sol:=y(x)=exp(m*x); > subs(sol,de); > simplify(%); sol := y(x) = exp(m x) / 3 \ / 2 \ |d | |d | /d \ |--- exp(m x)| + 3 |--- exp(m x)| + 2 |-- exp(m x)| = 0 | 3 | | 2 | \dx / \dx / \dx / 2 m exp(m x) (m + 3 m + 2) = 0 The polynomial has the same degree as the order of the differential

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equation. The polynomial equation is called the characteristic equation. We let Maple solve the cubic equation. > char_eq:=%/exp(m*x); > solve(%,{m}); 2 char_eq := m (m + 3 m + 2) = 0 {m = 0}, {m = -1}, {m = -2} The general solution (which includes all three constants of integration) is > sol:=y(x)=c1*exp(0*x)+c2*exp(-1*x)+c3*exp(-2*x); sol := y(x) = c1 + c2 exp(-x) + c3 exp(-2 x) Type: Linear homogeneous(rhs = 0) Cauchy-Euler equation. > de:=3*x^2*diff(y(x),x\$2)+11*x*diff(y(x),x)-3*y(x)=0; / 2 \ 2 |d | /d \ de := 3 x |--- y(x)| + 11 x |-- y(x)| - 3 y(x) = 0 | 2 | \dx / \dx / The coefficient of the second derivative is 3x^2, which has a second power of x; the coefficient of the coefficient of the first derivative is 11x, which has a first power of x; and the coefficient of the "zeroth derivative," y(x) has no powers of x. For Cauchy-Euler equations, the form of an explicit solution is always y(x) = x^m. > sol:=y(x)=x^m; > subs(sol,de); > simplify(%); m sol := y(x) = x / 2 \ 2 |d m| /d m\ m 3 x |--- x | + 11 x |-- x | - 3 x = 0 | 2 | \dx / \dx / m 2 m m 3 x m + 8 x m - 3 x = 0 The polynomial has the same degree as the order of the differential equation. The polynomial equation is called the indicial equation. We let
Maple solve the quadratic equation. > %/x^m; > ind_eq:=expand(%); > solve(%,{m}); m 2 m m 3 x m + 8 x m - 3 x ----------------------- = 0 m x 2 ind_eq := 3 m + 8 m - 3 = 0 {m = 1/3}, {m = -3} The general solution (which includes both constants of integration) is > sol:=y(x)=c1*x^(1/3)+c2*x^(-3); (1/3) c2 sol := y(x) = c1 x + ---- 3 x If there are initial values, so that we need to solve for the constants of the equation to get the unique solution, then we do that as in p. 15 #22.

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