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lindiff - Differential Equations and Linear Superposition...

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Differential Equations and Linear Superposition • Basic Idea: Provide solution in closed form • Like Integration, no general solutions in closed form Order of equation: highest derivative in equation e.g. is a 3rd order, non-linear equation. • First Order Equations: (separable, exact, linear, tricks) • A separable equation can be written: e.g. x 3 3 d dy x x d dy x 2 y ++ 0 = Ax (29 By x d dy dx == A x B y C + = x d dy 1 y 2 - 1 x 2 - + 0 = 1 y 2 - 1 x 2 - - = x 1 y 2 - y 1 x 2 - C 0 =
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• More Generally, an equation can be an exact differential: if the left side above is the exact differential of some u(x,y) then the solution is: • A necessary and sufficient condition for this is: e.g. is exact. We can then partially integrate A(x,y) in x and B(x,y) in y to guess a from for u(x,y): • For first order equations, it is always possible to find a factor λ (x,y) which makes the equation exact. Unfortunately, there is no general technique to find λ (x,y). .. Axy , (29 dx B x y , dy + 0 = uxy , C = y , x Bxy , = x 2 2 y + 2 x - y 2 + y = x 2 2 y + x 3 3 2 xy C y ++ = 2 2 - 2 y 3 3 - Cx + = , x 3 y 3 - 3 2 C =
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• Consider the general linear first-order equation: Lets try to find an integrating factor for this equation. .. We want to be exact. Equating the partial derivatives: This new equation for lambda is separable with solution: So we can always find an integrating factor for a first order linear differential equation, provided that we can integrate f(x). Then we can solve the original equation since it is now exact. e.g. The integrating factor is: So we get the exact equation: Integrating both sides: x d dy fx (29 y + gx = λ xy , λ x dy λ x ygx - dx + 0 = y Axy , x Bxy , = x d d λ x ⇒λ x = λ x e x d = 1 x + y + e x = y y x 1 + x + e x x = λ x e x d xe x == x y e x yx 1 + + e 2 x = x y e 2 x 2 Cy + e x 2 x Ce x - x +
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• Why do we call the equation below linear ? Consider a set of solutions: Any linear combination of the above solutions is another valid solution of the original equation. If we set: we get: which is clearly satisfied if y 1 and y 2 are indeed solutions. • Linear equations are satisfied by any linear superposition of solutions. We divide the set of solutions into a set of linearly independent solutions satis- fying the linear operator, and a particular solution satisfying the forcing func- tion g(x). In general, it can be shown that over a continuous interval, an equation of order k will have k linearly independent solutions to the homogenous equation (the linear operator), and one or more particular solutions satisfying the gen- eral (inhomogeneous) equation. A general solution is the superposition of a linear combination of homogenous solutions and a particular solution. • A wide variety of equations of interest to an electrical engineer are in fact linear. ..
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lindiff - Differential Equations and Linear Superposition...

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