Question #92902
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=ydy/dx=y'

d2y/dx2=yd^2y/dx^2=y''

(1+x)2y+(1+x)y+y=4cos(ln(1+x))(1+x)² y'' + (1+x) y' + y = 4 \cos (\ln(1+x))

Let's introduce the change of variable:

1+x=et1+x=e^t

t=ln(1+x)t=\ln(1+x)

dt/dx=1/(1+x)=etdt/dx=1/(1+x)=e^{-t}

y˙=dy/dt\dot {y}=dy/dt

y¨=d2y/dt2\ddot {y}=d^2y/dt^2

y=y˙ety'=\dot{y} e^{-t}

y=(y¨y˙)e2ty''=(\ddot{y} -\dot{y})e^{-2t}

So,

y¨+y˙+y=4cost\ddot{y}+\dot{y}+y=4\cos t

We obtained nonhomogeneous linear differential equation of the second order. The general solution of this nonhomogeneous equation is the sum of the general solution y0y_0 of the related homogeneous equation and a particular solution y1y_1 of the nonhomogeneous equation: y=y0+y1y=y_0+y_1.

Related homogeneous differential equation:

y¨+y˙+y=0\ddot{y}+\dot{y}+y=0

Its characteristic polynomial is r2+r+1r^2+r+1. Discriminant D=12411=3D=1^2-4 \cdot 1 \cdot 1=-3. So, the characteristic polynomial has two complex conjugate roots 1±i321=1/2±i3/2\frac{-1±i \sqrt{3}}{2 \cdot 1}=-1/2±i \sqrt{3}/2 and the general solution of homogeneous differential equation is y0=et/2(C1cos(3/2t)+C2sin(3/2t))y_0=e^{-t/2}(C_1 \cos (\sqrt{3}/2 \cdot t) + C_2 \sin (\sqrt{3}/2 \cdot t)).

Particular solution of the nonhomogeneous equation can be written in the form y1=Acost+Bsinty_1=A \cos t+B \sin t, where A and B - undetermined coefficients.

y1˙=Asint+Bcost\dot{y_1}=-A \sin t+B \cos t

y1¨=AcostBsint\ddot{y_1}=-A \cos t - B \sin t

Let's substitute these values into y¨+y˙+y=4cost\ddot{y}+\dot{y}+y=4\cos t:

AcostBsintAsint+Bcost+Acost+Bsint=4cost-A \cos t - B \sin t - A \sin t+B \cos t + A \cos t + B \sin t = 4 \cos t

BcostAsint=4costB \cos t - A \sin t = 4 \cos t

{A=0B=4\begin{cases} A=0 \\ B=4 \end{cases}

y1=4sinty_1=4 \sin t

So, y=y0+y1=et/2(C1cos(3/2t)+C2sin(3/2t))+4sinty=y_0+y_1=e^{-t/2}(C_1 \cos (\sqrt{3}/2 \cdot t)+ C_2 \sin (\sqrt{3}/2 \cdot t))+4 \sin t

The final answer can be obtained after the return to the variable x:

y=eln(1+x)2(C1cos(3/2ln(1+x))+C2sin(3/2ln(1+x)))+4sin(ln(1+x))y=e^{-\frac{\ln(1+x)}{2}}(C_1 \cos (\sqrt{3}/2 \ln(1+x) )+ C_2 \sin(\sqrt{3}/2 \ln(1+x)))+4 \sin (\ln(1+x))


Answer: y=eln(1+x)2(C1cos(3/2ln(1+x))+C2sin(3/2ln(1+x)))+4sin(ln(1+x))y=e^{-\frac{\ln(1+x)}{2}}(C_1 \cos (\sqrt{3}/2 \ln(1+x) )+ C_2 \sin(\sqrt{3}/2 \ln(1+x)))+4 \sin (\ln(1+x))


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