Question

d²y/dx² +2 dy/dx+y=e^-t lnt

Answer 1

$$solution$$

Consider the differential equation

$$y''+2y'+y=e^{-t}ln\ t$$

The characteristic equation is

$$m^2+2m+1=0\\ (m+1)^2=0\\ m+1=0\\ m=-1,-1$$

So, the complimentary solution of the differential equation is

$$y_c(t)=(c_1+c_2t)e^{-t}\\ y_c(t)=c_1e^{-t}+c_2te^{-t}\\$$

Let $y_1(t)=e^{-t}, y_2(t)=te^{-t}$

The Wronskian of these two functions is

$w(y_1(t), y_2(t))=\begin{pmatrix} e^{-t}& te^{-t} \\ -e^{-t} & e^{-t}-te^{-t} \\ \end{pmatrix}\\ \implies (e^{-2t}-te^{-2t})-(-te^{-2t})\\ \implies e^{-2t}$

And compare the given equation with $y''+p(t)y^t+q(t)y=g(t)$ then $p(t)=2,\ q(t)=1, \ g(t)=e^t\ ln\ t$

Now the particular solution is $yp(t)=y_1u_1+y_2u_2$ where

$u_1=-\intop\frac{y_2(t)g(t)}{w(y_1(t),y_2(t))}\delta t,\ u_2=\intop\frac{y_1(t)g(t)}{w(y_1(t),y_2(t))}\delta t,\\ u_1=-\intop\frac{(te^{-t})(e^{-t}ln\ t)}{e^{-2t}}\delta t,\ u_2=\intop\frac{(e^{-t})(e^{-t}ln\ t)}{e^{-2t}}\delta t,\\ u_1=-\intop(t\ ln\ t)\delta t,\ u_2=\intop(ln\ t)\delta t\\ u_1=-\frac{t^2}{4}(2\ ln\ t-1),\ u_2=t(ln\ t-1)$

Substitute $u_1, u_2$ in the particular solution $yp(t)=y_1u_1+y_2u_2$

Then is

$yp(t)=(e^{-t})(-\frac{t^2}{4}(2\ ln\ t-1))+(te^{-t})(t(ln\ t-1))\\ \implies (-\frac{t^2}{4}(2\ ln\ t-1))+(ln\ t-1)t^2e^{-t})\\ \implies (-\frac{ln\ t}{2}+\frac14+ln\ t-1)t^2e^{-t}\\ \implies (\frac{ln\ t}{2}-\frac34)t^2e^{-t}\\$

Now, the general solution is

$y(t)=y_p(t)+y_c(t)$

$\therefore$ The required solution is $y(t)=c_1e^{-t}+c_2te^{-t}+(\frac{ln\ t}{2}-\frac34)t^2e^{-t}$