Answer to Question #122600 in Differential Equations for Jse

Question

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation and a particular solution of the given nonhomogeneous equation.

y'' − 25y = 2; y1 = e^−5x

y1(x) = _____
y2(x) = ______

Answer 1

Solution:

y'' − 25y = 2;

y₁ = e^-5x 

y'' − 25y =0

y₂(x)=u(x)y₁(x)

y=ue^-5x

y'=u'e^-5x-5ue^-5x

y''=u''e^-5x-10u'e^-5x+25ue^-5x

y'' − 25y=u''e^-5x-10u'e^-5x+25ue^-5x-25ue^-5x=0

u''e^-5x-10u'e^-5x=0 | e^5x

u''-10u'=0

t=u'

t'-10t=0

dt/dx=10t

dt/t=10dx,

ln|t|=10x+C₁

t=Ce^10x

Let C=1.

y=e^10xe^-5x=e^5x

y₂(x)=e^5x

y_p(x)=A

A''=0

-25A=2=>A=-2/25

y_p(x)=-2/25

Answer:

y₂(x)=e^5x

^y_p(x)=-2/25