Solve:
(1+x)² d²y/dx² + (1+x) dy/dx + y = 4 cos ln(1+x)
1
Expert's answer
2019-08-23T10:11:29-0400
dy/dx=y′
d2y/dx2=y′′
(1+x)2y′′+(1+x)y′+y=4cos(ln(1+x))
Let's introduce the change of variable:
1+x=et
t=ln(1+x)
dt/dx=1/(1+x)=e−t
y˙=dy/dt
y¨=d2y/dt2
y′=y˙e−t
y′′=(y¨−y˙)e−2t
So,
y¨+y˙+y=4cost
We obtained nonhomogeneous linear differential equation of the second order. The general solution of this nonhomogeneous equation is the sum of the general solution y0 of the related homogeneous equation and a particular solution y1 of the nonhomogeneous equation: y=y0+y1.
Related homogeneous differential equation:
y¨+y˙+y=0
Its characteristic polynomial is r2+r+1. Discriminant D=12−4⋅1⋅1=−3. So, the characteristic polynomial has two complex conjugate roots 2⋅1−1±i3=−1/2±i3/2 and the general solution of homogeneous differential equation is y0=e−t/2(C1cos(3/2⋅t)+C2sin(3/2⋅t)).
Particular solution of the nonhomogeneous equation can be written in the form y1=Acost+Bsint, where A and B - undetermined coefficients.
y1˙=−Asint+Bcost
y1¨=−Acost−Bsint
Let's substitute these values into y¨+y˙+y=4cost:
−Acost−Bsint−Asint+Bcost+Acost+Bsint=4cost
Bcost−Asint=4cost
{A=0B=4
y1=4sint
So, y=y0+y1=e−t/2(C1cos(3/2⋅t)+C2sin(3/2⋅t))+4sint
The final answer can be obtained after the return to the variable x:
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