1 The Analytic Solution of a First-Order ODE

The ODE below has the following analytic solution,

y′ + 0.5y = 4e0.8x, y(0) = 2→ y =4

1.3(e0.8x − e−0.5x) + 2e−0.5x (1)

It is in the following form, where y′ + p(x)y = q(x) such that p(x) = 0.5 and q(x) = 4e0.8x. Hence, anintegrating factor could be used to evaluate the solution as follows.

USING AN INTEGRATING FACTOR TO SOLVE A LINEAR ODE:

If a first-order ODE can be written in the normal linear form,

y′ + p(x)y = q(x)

the ODE can be solved using an integrating factor µ(x) = eR

p(x)dx,

µ(x) [y′ + p(x)y] = µ(x)q(x)

(µ(x)y)′ = µ(x)q(x)

µ(x)y =

∫µ(x)q(x)dt+ C

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

Since µ(x) = eR

0.5 dx = e0.5x,

e0.5xy = 4

∫e1.3x dx+ C = 4

(e1.3x

1.3

)+ C =

4

1.3e1.3x + C

y =4

1.3e0.8x + Ce−0.5x

(2)

Since y(0) = 2,

2 =4

1.3+ C → C = 2− 4

1.3(3)

Hence,

y =4

1.3e0.8x +

(2− 4

1.3

)e−0.5x =

4

1.3

(e0.8x − e−0.5x

)+ 2e−0.5x (4)