Download Solving Linear First-Order Differential Equations using an Integrating Factor - Prof. Brya and more Study notes Linear Algebra in PDF only on Docsity!
1.5 Linear First-Order Differential Equations Our second method of solving a differential equation is… using an integrating factor. A linear first-order differential equation has the form P ( x ) y Q ( x ) dx dy . Notice that the equation is linear in the dependent variable y (virtually anything is allowed in terms of the independent variable x ). The key is to recognize that the only terms involving y are the derivative term y ´ and y times some function of x. This is a special form of differential equation having a distinct solution method. Due to the subtleties of specific examples, we will always use the method of solution instead of attempting to use a formula to solve these types of equations. This method examines the left hand side and recognizes that it can be ‘forced’ to be the derivative of a product assuming that P ( x ) comes from the chain rule for some function ^ ( x ). For every linear first-order differential equation, we must have ( x ) e P ( x ) dx. We refer to (^) ( x ) e P ( x ) dx as an integrating factor since multiplying both sides of the differential equation by (^) ( x ) e P ( x ) dx gives ( ) ( ) () () () ( ) e P xe y e Q x dx dy Pxdx dx^ d e^ y Pxdx Px dx Px dx
This allows us to rewrite the left side of the differential equation as a derivative: ( ) () (^ ) dx^ d^ e y e Q^ x Pxdx Pxdx ^ ^ ^ Therefore, integrating both sides with respect to x gives
e^ P^^ (^ x^ )^ dx y e P^ ( x ) dxQ ( x ) dx
Solving for y results in y e ^ P^ (^ x^ )^ dx e ^ P ( x ) dxQ ( x ) dx It is tempting to offer a formula representing y (we could) but there are subtleties regarding the constant of integration corresponding to given initial values. It is advised that you learn to rely on the method itself rather than a formulated solution.
Consider the following examples: Example: First, verify that x^3^ x^3 y xe ^ Ce satisfies the differential equation 2 3 3 x y ex dx dy (^) We have 3 3 3 y ' e ^ x^ 3 x^2 e x 3 xCe x so substitution into the differential equation gives 3 3 3 3 3 3 ' 3 2 3 3 3 2 3 2 ( ) x x x x x x e y x y e xe x Ce x xe Ce therefore x^3^ x^3 y xe ^ Ce is a solution to the differential equation. Now let’s solve the differential equation 2 3 3 x y ex dx dy (^) (Here P ( x ) 3 x^2 and 3 Q ( x ) e x .) using an integrating factor. In this case, we determine the integrating factor to be 32 3 ( ) x dx x x e e.
Multiplying both sides of the differential equation by^ ^ yields
3 2 3 3 3 x (^) 3 xex y exe x dx dy e allowing us to rewrite the equation as
3 e y dx d (^) x . Integrating both sides with respect to x gives ex^ y x C 3 . Solving for y , we obtain the afore-mentioned solution x^3 x^3 y xe ^ Ce . The integral obtained on the right side was pretty simple. Consider how a small change can effect the integration: Example: Solve the differential equation x y ex^ x dx dy (^) 2 3 (^3) Here P ( x ) 3 x^2 and Q x e x x 3 ( ). The integrating factor in this case is again 32 3 ( ) x xdx
x e e. Multiplying both sides by^
yields x (^) xex y e x dx dy e 3 2 3 3 allowing us to rewrite the equation as
3 e y dx d (^) x .
Application Our primary example from this section is a mixture problem that we can create the resulting differential equation from scratch. Consider a tank (reservoir, vat, swimming pool, lemonade dispenser, whatever) containing a solution – a mixture of a solute and solvent – such as salt dissolved in water. There is inflow and outflow and we seek to determine a formula for the amount x ( t ) of salt (solute) in the tank at time t , given the amount x (0) = x 0 at time t = 0. We note the following relationships: r i = rate flowing into the tank c i = concentration of solute per unit volume coming into the tank V ( t ) = volume of tank at time t r o = rate flowing out of the tank c o = concentration of solute per unit volume coming out of the tank = () ( ) V t x t The differential equation thus describes the change in amount of solute with respect to time t. Therefore, we approximate the change in solute by x ri ci roco t r i Dividing by t and passing to the limit as t 0 , we obtain Volume: V ( t ) x ( t ) ( ) ( ) xt V t r rc rc r c dt dx (^) o (^) i i o o i i r o This results in the linear differential equation i i o (^) xt rc V t r dt dx () ( ) Example: Set up and solve the differential equation for a 150 gallon tank that initially ( t = 0) contains 60 pounds of salt dissolved in 90 gal of water. Brine (salt water) containing 2 lb/gal of salt flows into the tank at a rate of 5 gal/min and the well-stirred mixture flows out of the tank at a rate of 3 gal/min. Find the amount of salt x ( t ) in the tank t minutes later. How much salt is in the tank when it is full? Differential Equation: ( ) 90 2 3 10 xt dt t dx Standard Linear First-Order Form: 10 90 2 3 x dt t dx In this case, the integrating factor is 2 ln(^902 )^32 3 90 2 3 ( t ) e e ( 90 2 t ) t dt t
.
Going directly to the resulting integral equation, we have t x t dt t^2 C 5 2 3 2 3
Solving for x ( t ) gives 2 3
t
C
xt t
Imposing the initial condition that x (0) = 60 pounds of salt gives 2 3 2 3 120 (^90 ) 90 60 180 C C . The solution is 2 3 90 2 90 ( ) 2 ( 90 2 ) (^120) t xt t The tank is full when t = 30 minutes, so there is
- 2 150 90 ( 30 ) 300 120 2 3 ^ x ^ pounds of salt in the tank.
