Answer to Question #147905 in Differential Equations for ifad

Question

(x^2D^2-3xD + 1)y= x^-1 [1 + logx sin (logx)]

Answer 1

$$Solution$$

Given the equation

$$(x^2D^2-3xD + 1)y= x^{-1} [1 + logx sin (logx)]$$

which is a homogeneous linear equation. Let $x=e^z$, that is $z=log\ x$, then the equation becomes;

$(D_1(D_1-1)-3D_1+1)y=e^z(1+z\sin z)\\ \implies (D_1^2-4D_1+1)y=e^z(1+z\sin z)$

Hence the auxiliary equation is

$(D_1^2-4D_1+1)y=0\\ \implies D_1=\frac{4 \pm \sqrt{12}}{2}=2 \pm \sqrt{3}$

Hence $C.F=C_1e^{(2+\sqrt{3})z}+C_2e^{(2-\sqrt{3})z}$

$C.F=C_1e^{(2+\sqrt{3})log x}+C_2e^{(2-\sqrt{3})log x}\\ \implies C_1x^{2+\sqrt{3}}+C_2x^{2-\sqrt{3}}\\ \implies x^2(C_1x^{\sqrt{3}}+C_2x^{-\sqrt{3}})$

Now,

$$P.I=\frac{1}{D^2_1-4D_1+1}e^{-z}(z\ sin z)+e^{-z}$$

For $e^{-z}(z\ sin z)$

$P.I_1=\frac{1}{D^2_1-4D_1+1}e^{-z}(z\ sin z)\\ \implies e^{-z}\frac{1}{(D_1-1)^2-4(D_1-1)+1}(z\ sin z)\\ \implies e^{-z}\frac{1}{(D_1^2-6D_1+6)}(z\ sin z)\\ \implies e^{-z}[z \frac{1}{D_1^2-6D_1+6}sin z-\frac{1}{(D_1^2-6D_1+6)^2}(2D_1-6)sin z]\\ \implies e^{-z}[z \frac{1}{D_1^2-6D_1+6}sin z-\frac{1}{(D_1^2-6D_1+6)^2}(2cos z-6sin z)]\\ \implies e^{-z}[z \frac{1}{5-6D_1}sin z-\frac{1}{(5-6D_1)^2}(6sin z-2cos z)]\\ \implies e^{-z}[z(5+6D_1) \frac{1}{25-36D_1^2}sin z-\frac{1}{25-60D_1+36D_1^2}(6sin z-2cos z)]\\ \implies e^{-z}[z(5+6D_1) \frac{1}{61}sin z-\frac{1}{-11-6-D_1}(6sin z-2cos z)]\\ \implies e^{-z}[\frac{z}{61}(5sin z +6cos z)-(60D_1-11)\frac{1}{3600D_1^2-121}(6sin z-2cos z)]\\ \implies e^{-z}[\frac{z}{61}(5sin z +6cos z)+(60D_1-11)\frac{1}{3721}(6sin z-2cos z)]\\ \implies e^{-z}[\frac{z}{61}(5sin z +6cos z)+\frac{1}{3721}(-360cos z-120sin z-66sinz-22cos z)]\\ \implies e^{-z}[\frac{z}{61}(5sin z +6cos z)+\frac{1}{3721}(-382cos z-54sin z)]\\ \implies x^{-1}[\frac{log x}{61}(5sin (log x) +6cos (log x))+\frac{1}{3721}(-382cos (log x)-54sin (log x))]\\$

$\therefore P.I_1= x^{-1}[\frac{log x}{61}(5sin (log x) +6cos (log x))+\frac{1}{3721}(-382cos (log x)-54sin (log x))]$

Now, for $e^{-z}$ where $D_1=-1$

$P.I_2=\frac{1}{D_1^2-4D_1+1}e^{-z}\\ P.I_2=\frac{1}{(1)^2-4(1)+1}e^{-z} \implies \frac{1}{(1-4+1)e^{-z}}=\frac{1}{6}x\\$

Hence the general solution is $y=C.F+P.I_2+P.I_1$

$y=x^2(C_1x^{\sqrt{3}}+C_2x^{-\sqrt{3}})+\frac{1}{6}x+x^{-1}[\frac{log x}{61}(5sin (log x) +6cos (log x))+\frac{1}{3721}(-382cos (log x)-54sin (log x))]$