Analytic Solution Of A First-order Ode (michael De Silva)

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The Analytic Solution of a First-Order ODE

The ODE below has the following analytic solution, y 0 + 0.5y = 4e0.8x , y(0) = 2 → y =

4 0.8x (e − e−0.5x ) + 2e−0.5x 1.3

(1)

It is in the following form, where y 0 + p(x)y = q(x) such that p(x) = 0.5 and q(x) = 4e0.8x . Hence, an integrating factor could be used to evaluate the solution as follows. U SING AN I NTEGRATING FACTOR TO SOLVE A L INEAR ODE: If a first-order ODE can be written in the normal linear form, y 0 + p(x)y = q(x) the ODE can be solved using an integrating factor µ(x) = e

R

p(x)dx

,

µ(x) [y 0 + p(x)y] = µ(x)q(x) (µ(x)y)0 = µ(x)q(x) Z µ(x)y =

µ(x)q(x)dt + C

Dividing through by µ(x), we have the general solution of the linear ODE. R

Since µ(x) = e 0.5x

e

0.5 dx

Z y=4

= e0.5x ,

1.3x

e

 dx + C = 4

e1.3x 1.3

 +C =

4 1.3x e +C 1.3

(2)

4 0.8x y= e + Ce−0.5x 1.3 Since y(0) = 2, 2=

4 4 +C →C =2− 1.3 1.3

(3)

Hence,  
4 0.8x 4 4 y= e + 2− e−0.5x = e0.8x − e−0.5x + 2e−0.5x 1.3 1.3 1.3

(4)