Linear Differential Equations | lesson 2

Linear equations - use of integrating factor

Consider the equation dy/dx + 5y = e²ˣ
This is clearly an equation of the first order , but different from those we have dealt with so far. In fact, none of our previous methods could be used to solve this one, so we have to find a further method of attack.

The above equation can be express in the form dy/dx + Py = Q , where P and Q are functions of x ( or constants ).
This equation is called a linear equation of the first order and to solve any such equation, we multiply both sides by an integrating factor which is always e^∫pdx
This converts the LHS into the derivative of the product of y and the integrating factor.

So to solve dy/dx + 5y = e²ˣ
where P = 5, therefore ∫pdx = 5x and the integrating factor is therefore e⁵ˣ
Note : In determining ∫pdx, we do not include a constant of integration. This omission is purely for convenience, for a constant of integration here would in practice give a constant factor on both sides of the equation, which would cubsequently cancel. This is one of the rare occasions when we do not write down the constant of integration.

So to solve dy/dx + 5y = e²ˣ we will multiply each term by the integrating factor
e⁵ˣ(dy/dx) + 5ye⁵ˣ=e⁷ˣ
The LHS can be express as d/dx(ye⁵ˣ)
Hence:
d/dx(ye⁵ˣ) = e⁷ˣ
By integrating both sides, we have
ye⁵ˣ= ⅐e⁷ˣ +c
and y = ⅐e²ˣ+ce⁻⁵ˣ

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