Answer to Question #92902 in Differential Equations for Subhasis Padhy

Question

Solve:
(1+x)² d²y/dx² + (1+x) dy/dx + y = 4 cos ln(1+x)

Answer 1

$dy/dx=y'$

$d^2y/dx^2=y''$

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

Let's introduce the change of variable:

$1+x=e^t$

$t=\ln(1+x)$

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

$\dot {y}=dy/dt$

$\ddot {y}=d^2y/dt^2$

$y'=\dot{y} e^{-t}$

$y''=(\ddot{y} -\dot{y})e^{-2t}$

So,

$\ddot{y}+\dot{y}+y=4\cos t$

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

Related homogeneous differential equation:

$\ddot{y}+\dot{y}+y=0$

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

Particular solution of the nonhomogeneous equation can be written in the form $y_1=A \cos t+B \sin t$, where A and B - undetermined coefficients.

$\dot{y_1}=-A \sin t+B \cos t$

$\ddot{y_1}=-A \cos t - B \sin t$

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

$-A \cos t - B \sin t - A \sin t+B \cos t + A \cos t + B \sin t = 4 \cos t$

$B \cos t - A \sin t = 4 \cos t$

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

$y_1=4 \sin t$

So, $y=y_0+y_1=e^{-t/2}(C_1 \cos (\sqrt{3}/2 \cdot t)+ C_2 \sin (\sqrt{3}/2 \cdot t))+4 \sin t$

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

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

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