Section 2.4 Linear Equations

Transcription

1 Section 2.4 Linear Equations Key Terms: Linear equation Homogeneous linear equation Nonhomogeneous (inhomogeneous) linear equation Integrating factor General solution Variation of parameters

2 A first-order linear equation is one of the form x' = a(t)x + f (t) or equivalent x ' a(t)x = f (t) where functions a(t) and f(t) are called the coefficients. If f(t) = 0, the equation has the form x ' = a(t)x or equivalently x ' a(t)x = 0 and the linear equation in this case is said to be homogeneous. Otherwise it is called nonhomogeneous (inhomogeneous). In this case the DE is separable. We will sometimes consider equations of the more general form b(t)x ' = c(t)x + g(t) which can be put into the form x ' a(t)x = f(t) by dividing by b(t) 0. The standard form of a first order linear equation is x ' a(t)x = f(t). We will use this form to develop a procedure for the general solution of first order linear DEs.

3 Examples: Homogeneous linear DE We use different dependent and independent variables at times.

4 Solving homogeneous linear DEs For homogeneous DE x ' = a(t)x we can separate the variables and integrate: Exponentiating, we get The constant e C is positive. We often replace it with the symbol C. We will allow C to be positive, negative, or zero, so that we can get rid of the absolute value. Hence the general solution of a first order homogeneous linear DE is x(t) = Ce a(t) Example: Solve x ' = cos(2t)x. Instead of memorizing the formula for the solution, just recognize that the DE is separable and proceed. 1 1 sin(2t) 2 2 dx 1 sin(2t)+c = cos(2t) ln x = sin(2t) + C x = e x = Ce x 2

5 Solving nonhomogeneous linear DEs First put the nonhomogeneous DE into standard form x ' a(t)x = f (t). The left side, x'(t) - a(t)x looks like a "piece of a product rule". Recall that d ( u(t)x(t) ) = u(t)x'(t) +u'(t)x(t) We would like to find a function u(t) so that when we multiply both sides of x ' a(t)x = f(t) the left side is the derivative of the product u(t)x(t). That is, (u(t)x(t))' = u(t)[x' (t) a(t)x(t)]. If we can find such a function u(t) it is called an integrating factor. In such a case we would have (u(t)x(t))' =u(t)[x' (t) a(t)x(t)] = u(t)f(t). Integrating both sides we get utxt ( ) ( ) = utft ( ) ( ) +C Dividing by u(t) we have general solution of a first order linear nonhomogeneous DE 1 C x t = u t f t + u(t) ( ) ( ) ( ) u(t)

6 Thus, the key to the method is finding an integrating factor, a function u(t) that satisfies equation, (u(t)x(t))' = u(t)[x'(t) a(t)x(t)]. If we expand both sides of this expression we have (dropping the t notation) ux' + u'x = ux' aux. Subtracting ux' from each side, this becomes u'x = aux, which will be satisfied if Solving this DE we get - a(t) u(t) = e u' = au. our integrating factor. (Note that we do not need to include a constant of integration here, rather we just need one particular solution.)

7 Summary: To solve the nonhomogeneous first order linear DE in standard form x ' a(t)x = f(t) do the following: Compute the integrating factor: - a(t) u(t) = e Multiply both sides of the DE by integrating factor u(t). The result is that the DE can now be written in the form d ( utxt ( ) ( )) = utft ( ) ( ) Integrate both sides to obtain utxt ( ) ( ) = utft ( ) ( ) +C Finally solve for x(t). 1 C x t = u t f t + u(t) ( ) ( ) ( ) u(t) Note: if the DE has the form x ' + a(t)x = f(t) then u(t) = e a(t)

8 Example: Solve 5t x' = -3x + e Put into standard form: x' ( 3) x = e 5t a(t) = -3, f(t) = e 5t - a(t) 3 Compute the integrating factor: u(t) = e = e Multiply both sides of the DE by integrating factor u(t). Now the DE is d ( ) d u t x t = u t f t e x t = e e = e ( ( ) ( )) ( ) ( ) ( ) Integrate both sides to obtain 3t 8t 1 8t e x t = e = e + C 8 ( ) 3t 3t 5t 8t Finally solve for x(t). 1 e 8t +C 1 x ( t 8 ) = = e + Ce 3t e 8 5t -3t

9 dx Example: Solve - 2x = 4 - t This is in standard form so a(t) = 2 and f(t) = 4 t. Compute the integrating factor: u(t) = e = e - at () 2 Multiply both sides of the DE by integrating factor u(t). Now the DE is d ( ) d u t x t = u t f t e x t = e (4 t) = 4e te ( ( ) ( )) ( ) ( ) ( ) 2 t 2 t 2 t 2 t Integrate both sides to obtain -2t ( ) -2t -2t e x t = ( 4e -te ) +C Need integration by parts. We get -2t -2t 1-2t 1-2t e x =-2e + te + e +C 2 4 Now solve for x: 1 1-2t t x = -2 + t + + Ce = + t + Ce t dx This solution has a special structure: Ce is the general solution of - 2x = dx + t is a particular solution of - 2x = 4 - t 4 2 x h x p

10 Alternate approach: There is another method of solving linear equations that you might find easier to remember and use. (It still requires two integrations.) For the first order linear DE y' = a(t)y + f(t) we solve the associated homogeneous DE y h ' = a(t)y h. Separating the variables and integrating we have y =e h a(t) Notice that because of its exponential form the function y h is never equal to zero. Hence we can safely divide by it. We now seek a solution y to the nonhomogeneous problem that has the form y = v(t)y h, where v(t) is a function to be determined. We substitute v(t)y h into the original DE. (vy h )ꞌ = a(vy h ) + f or vy' h + v' y h = avy h + f Rearrange to get v(y' h - ay h ) + v' y h = f and recall that y h ' = ay h so the result is v'y h = f or equivalently solution is y = vy h. f v' = y This method is called variation of parameters. h. We integrate to find function v. Then the

11 Example: Solve IVP y' = -2ty+2t 3 by variation of parameters. The associated homogeneous DE is y h ' = -2ty h. Separating the variables then integrating we get dy 2 h 2 h -t Here y h is the = -2t ln y = -t y h = e solution of the yh homogeneous DE. We omitted the constant of integration C since we need only one solution to the homogeneous DE. From the method of variation of parameter we now seek a solution to the nonhomogeneous DE of the form y = vy h. Substituting this expression into the original equation we obtain a DE to solve for function v (after simplification); f 2t v' = = = 2t e t yh e -t y p y h We integrate by parts to find v, this time attaching a constant of integration. 2 t2 v = (t - 1)e + C 2 2 Thus the general solution of the DE is y = vy h = t -t -t (t - 1)e + C e = (t - 1) + Ce It is easy to check that t is a particular solution of the nonhomogeneous DE From MATLAB >> int(sym('2*t^3*exp(t^2)')) ans = exp(t^2)*(t^2-1)

12 Structure of the solution: A first order linear DE x' = a(t)x + f (t) has a particular structure to its general solution. Suppose that y p is a particular solution to the nonhomogeneous equation y = a(t)y + f (t), and that y h is a particular solution to the associated homogeneous equation. Then every solution to the inhomogeneous equation is of the form where C is an arbitrary constant. y(t) = y p (t) + Cy h (t), Note by particular solution we mean we omit the constant of integration. This form was evident in the examples we solved by both techniques.