First Order Linear D.E.

Example: Solve the d.e. y’ + y = e^2x

with the initial condition: y(0) = 1

Strategy: Use an integrating factor (IF).

The general form of a linear, first order, d.e. is y’ + p(x) y = q(x)

In this example: p(x) = 1,

So the integrating factor is IF = e ^{∫ dx} = e^x

Next Step: Multiply each term of the original d.e. by the integrating factor. The result

is

e^x y’ + e^x y = e^3x

Next Step: Factor terms on the left hand side as follows:

d/dx [ e^x y ] = e^3x

NOTE: If you calculated the correct IF, then you should have arrived with an

expression on the left hand side of the d.e. that you can integrate.

(This result serves as a check on your IF.)

e^x y = ∫ e^3x dx = (1/3) e^3x + C

y(x) = (1/3) e^2x + C e^-x where C is a constant to be determined

from a specified initial condition such as

y(0) = 1

In this case 1 = 1/3 + C, So C = 2/3

And y(x) = (1/3) e^2x + (2/3) e^-x (result)

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