dx2D2y−24dxdy+144=0
the differential equation is of the form
y′′+py′+qy=s
p=-24
q=0
s=-144
It is linear inhomogeneous
second -order differential equation with constant coefficients.
the equation has an easy solution
first we should find the roots of the characteristic equation
q+(k2+kp)=0
k2+kp=0
this is simple equation
the roots of this equation are
k1=0k2=24
as there are two roots of characteristic equation ,
and the roots are not complex , then
solving the corresponding differential equation looks as follows
⟹
y(x)=C1ek1x+C2ek2x
y(x)=C1+C2e24x
we get asolution for the corresponding homogeneous equation
Now we should solve the inhomogeneous equation
y''+py'+qy=s
use variation of parameters method
Suppose that C1 and C2 - it is function of x
tjhe general solution is
y(x)=C1(x)+C2(x)e24x
where C1(x) and C2(x)
by the method of variation of parameters , we find the solution from the system:
y1(x)dxdC1(x)+y2(x)dx2dC2(x)=0
dxdC1(x)dxdy1(x)+dxdC2(x)dxdy2(x)=f(x)
where
y1(x) and y2(x) - linearly independent particular solutions of Linear Ordinary Differential Equations,
y1(x)=1 (C1=1 , C2=0),
y2(x)=e(24x) (C1=0 , C2=1)
the free term f=s or
f(x)=−144
so the system has the form
e24xdxdC2(x)+dxdC1(x)=0dxd1dxdC1(x)+dxdC2(x)dxde24x=−144ore24xdxdC2(x)+dxdC1(x)=024e24xdxdC2(x)=−144
solve the system :
dxdC1(x)=6dxdC2(x)=−6e−24x
it is the simple equation , solve these equation
C1(x)=C3+∫6dxC2(x)=C4+∫(−6e−24x)dx
or
C1(x)=C3+6xC2(x)=C4+4e−24
substitute found C1(x) and C2(x) to
y(x)=C1(x)+C2(x)e24x
the answer is
y(x)=C3+C4e24x+6x+41
where C3 and C4 are constant and also41
so final answer is
y(x)=C1+C2e24x+6xanswer
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