Differential-Equations Second Order Linear Differential Equations, guaranteed and verified 100% Pass

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Second Order Linear Differential Equations


A second order linear differential equation has the form:

𝐴(𝑥)𝑦" + 𝐵(𝑥)𝑦′ + 𝐶(𝑥)𝑦 = 𝐹(𝑥).
If 𝐹 (𝑥 ) = 0, we say the equation is homogeneous.



Ex. 𝑥 2 𝑦" + 2𝑥𝑦 ′ + 3𝑦 = cos 𝑥 is a non-homogeneous linear equation.
𝑥 2 𝑦" + 2𝑥𝑦 ′ + 3𝑦 = 0 is the associated homogeneous equation.


Notice that if 𝐴(𝑥)𝑦" + 𝐵(𝑥)𝑦′ + 𝐶(𝑥)𝑦 = 𝐹(𝑥) and 𝐴(𝑥) ≠ 0, then we
can divide the equation by 𝐴(𝑥) and get an equation of the form:

𝑦" + 𝑝(𝑥)𝑦′ + 𝑞(𝑥)𝑦 = 𝑓(𝑥).
Here, the associated homogeneous equation is:

𝑦" + 𝑝(𝑥)𝑦′ + 𝑞(𝑥)𝑦 = 0.


Notice that if 𝑦1 and 𝑦2 are solutions to a homogeneous equation then so is
𝑐1 𝑦1 + 𝑐2 𝑦2 for any real numbers, 𝑐1 and 𝑐2 .
If 𝑦1 , 𝑦2 are solutions to:

𝑦" + 𝑝(𝑥)𝑦′ + 𝑞(𝑥)𝑦 = 0
then: 𝑦1 ′′ + 𝑝(𝑥)𝑦1 ′ + 𝑞(𝑥)𝑦1 = 0
𝑦2 ′′ + 𝑝(𝑥)𝑦2 ′ + 𝑞(𝑥)𝑦2 = 0.

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So we have:
(𝑐1 𝑦1 + 𝑐2 𝑦2 )′′ + 𝑝(𝑥 )(𝑐1 𝑦1 + 𝑐2 𝑦2 )′ + 𝑞 (𝑥 )(𝑐1 𝑦1 + 𝑐2 𝑦2 )
= (𝑐1 𝑦1′′ + 𝑐2 𝑦2′′ ) + 𝑝(𝑥 )(𝑐1 𝑦1′ + 𝑐2 𝑦2′ ) + 𝑞(𝑥)(𝑐1 𝑦1 + 𝑐2 𝑦2 )
= 𝑐1 (𝑦1′′ + 𝑝(𝑥 )𝑦1′ + 𝑞 (𝑥 )𝑦1 ) + 𝑐2 (𝑦2′′ + 𝑝(𝑥 )𝑦2′ + 𝑞(𝑥 )𝑦2 )
= 0.


Existence and Uniqueness Theorem: Suppose 𝑝(𝑥 ), 𝑞 (𝑥 ), and 𝑓(𝑥) are
continuous on an open interval, 𝐼, containing the point 𝑎. Then given any two
numbers, 𝑏0 and 𝑏1 , the equation:

𝑦" + 𝑝(𝑥)𝑦′ + 𝑞(𝑥)𝑦 = 𝑓(𝑥)
has a unique solution on the interval, 𝐼, that satisfies

𝑦(𝑎) = 𝑏0 , 𝑦 ′ (𝑎) = 𝑏1 .


Ex. Perform the following:
a) Verify that 𝑦1 = 𝑒 5𝑥 , 𝑦2 = 𝑥𝑒 5𝑥 are solutions to 𝑦" − 10𝑦′ + 25𝑦 = 0.
b) Find a unique solution for the initial conditions 𝑦(0) = 3, 𝑦 ′ (0) = 13.

a) 𝑦1 = 𝑒 5𝑥 𝑦2 = 𝑥𝑒 5𝑥
𝑦1′ = 5𝑒 5𝑥 𝑦2′ = 5𝑥𝑒 5𝑥 + 𝑒 5𝑥
𝑦1′′ = 25𝑒 5𝑥 𝑦2′′ = 25𝑥𝑒 5𝑥 + 5𝑒 5𝑥 + 5𝑒 5𝑥 = 25𝑥𝑒 5𝑥 + 10𝑒 5𝑥

𝑦1′′ − 10𝑦1′ + 25𝑦1 = 25𝑒 5𝑥 − 10(5𝑒 5𝑥 ) + 25(𝑒 5𝑥 )
= 25𝑒 5𝑥 − 50𝑥𝑒 5𝑥 + 25𝑒 5𝑥
=0