Question

Solve the differential equation y
′′
+
2
y
′
+
y
=
e
−
2
t
ln
t
by variation of parameters.

Answer 1

Answer:

$y(t)=c_1e^{-t}+c_2te^{-t}+\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}$

Step-by-step explanation:

Given the second-order differential equation. Solve by using variation of parameters.

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

(1) - Solve the DE as if it were homogeneous to find the homogeneous solution

$y''+2y'+y=e^{-t}\ln(t) \Longrightarrow y''+2y'+y=0\\\\\text{The characteristic equation} \rightarrow m^2+2m+1=0, \ \text{solve for m}\\\\m^2+2m+1=0\\\\\Longrightarrow (m+1)(m+1)=0\\\\\therefore \boxed{m=-1,-1}$

$\boxed{\left\begin{array}{ccc}\text{\underline{Solutions to Higher-order DE's:}}\\\\\text{Real,distinct roots} \rightarrow y=c_1e^{m_1t}+c_2e^{m_2t}+...+c_ne^{m_nt}\\\\ \text{Duplicate roots} \rightarrow y=c_1e^{mt}+c_2te^{mt}+...+c_nt^ne^{mt}\\\\ \text{Complex roots} \rightarrow y=c_1e^{\alpha t}\cos(\beta t)+c_2e^{\alpha t}\sin(\beta t)+... \ ;m=\alpha \pm \beta i\end{array}\right}$

Notice we have repeated/duplicate roots, form the homogeneous solution.

$\boxed{\boxed{y_h=c_1e^{-t}+c_2te^{-t}}}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now using the method of variation of parameters, please follow along very carefully.

$\boxed{\left\begin{array}{ccc}\text{\underline{Variation of Parameters Method(1 of 2):}}\\ \text{Given a DE in the form} \rightarrow ay''+by"+cy=g(t) \\ \text{1. Obtain the homogenous solution.} \\ \Rightarrow y_h=c_1y_1+c_2y_2+...+c_ny_n \\ \\ \text{2. Find the Wronskain Determinant.} \\ |W|= \left|\begin{array}{cccc}y_1 & y_2 & \dots & y_n \\y_1' & y_2' & \dots & y_n' \\\vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \dots & y_n^{(n-1)}\end{array}\right| \\ \\ \end{array}\right}$

$\boxed{\left\begin{array}{ccc}\text{\underline{Variation of Parameters Method(2 of 2):}}\\ \text{3. Find} \ W_1, \ W_2, \dots, \ W_n.\\ \\ \text{4. Find} \ u_1, \ u_2, \dots, \ u_n. \\ \Rightarrow u_n= \int\frac{W_n}{|W|} \\ \\ \text{5. Form the particular solution.} \\ \Rightarrow y_p=u_1y_1+u_2y_2+ \dots+ u_ny_n \\ \\ \text{6. Form the general solution.}\\ y_{gen.}=y_h+y_p\end{array}\right}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(2) - Finding the Wronksian determinant

$|W|= \left|\begin{array}{ccc}e^{-t}&te^{-t}\\-e^{-t}&e^{-t}-te^{-t}\end{array}\right|\\\\\Longrightarrow (e^{-t})(e^{-t}-te^{-t})-(te^{-t})(-e^{-t})\\\\\Longrightarrow (e^{-2t}-te^{-2t})-(-te^{-2t})\\\\\therefore \boxed{|W|=e^{-2t}}$

(3) - Finding W_1 and W_2

$W_1=\left|\begin{array}{ccc}0&y_2\\g(t)&y_2'\end{array}\right| \ \text{Recall:} \ g(t)=e^{-t} \ln(t)\\\\\Longrightarrow \left|\begin{array}{ccc}0&te^{-t}\\e^{-t} \ln(t)&e^{-t}-te^{-t}\end{array}\right|\\\\\Longrightarrow 0-(te^{-t})(e^{-t} \ln(t))\\\\\therefore \boxed{W_1=-t\ln(t)e^{-2t}}$

$W_2=\left|\begin{array}{ccc}y_1&0\\y_1'&g(t)\end{array}\right| \ \text{Recall:} \ g(t)=e^{-t} \ln(t)\\\\\Longrightarrow \left|\begin{array}{ccc}e^{-t}&0\\-e^(-t)&e^{-t} \ln(t)\end{array}\right|\\\\\Longrightarrow (e^{-t})(e^{-t} \ln(t))-0\\\\\therefore \boxed{W_2=\ln(t)e^{-2t}}$

(4) - Finding u_1 and u_2

$u_1=\int \frac{W_1}{|W|}; \text{Recall:} \ W_1=-t\ln(t)e^{-2t} \ \text{and} \ |W|=e^{-2t} \\\\\Longrightarrow \int\frac{-t\ln(t)e^{-2t}}{e^{-2t}} dt\\\\\Longrightarrow -\int t\ln(t)dt \ \text{(Apply integration by parts)}\\\\\\\boxed{\left\begin{array}{ccc}\text{\underline{Integration by Parts:}}\\\\uv-\int vdu\end{array}\right }\\\\\text{Let} \ u=\ln(t) \rightarrow du=\frac{1}{t}dt \\\\\text{an let} \ dv=tdt \rightarrow v=\frac{1}{2}t^2$

$\Longrightarrow -\Big[(\ln(t))(\frac{1}{2}t^2)-\int [(\frac{1}{2}t^2)(\frac{1}{t}dt)]\Big]\\\\\Longrightarrow -\Big[\frac{1}{2}t^2\ln(t)-\frac{1}{2}\int (t)dt\Big]\\\\\Longrightarrow -\Big[\frac{1}{2}t^2\ln(t)-\frac{1}{2}\cdot\frac{1}{2}t^2 \Big]\\\\\therefore \boxed{u_1=\frac{1}{4}t^2-\frac{1}{2}t^2\ln(t)}$

$u_2=\int \frac{W_2}{|W|}; \text{Recall:} \ W_2=\ln(t)e^{-2t} \ \text{and} \ |W|=e^{-2t} \\\\\Longrightarrow \int\frac{\ln(t)e^{-2t}}{e^{-2t}} dt\\\\\Longrightarrow \int \ln(t)dt \ \text{(Once again, apply integration by parts)}\\\\\text{Let} \ u=\ln(t) \rightarrow du=\frac{1}{t}dt \\\\\text{an let} \ dv=1dt \rightarrow v=t \\\\\Longrightarrow (\ln(t))(t)-\int[(t)(\frac{1}{t}dt )] \\\\\Longrightarrow t\ln(t)-\int 1dt\\\\\therefore \boxed{u_2=t \ln(t)-t}$

(5) - Form the particular solution

$y_p=u_1y_1+u_2y_2\\\\\Longrightarrow (\frac{1}{4}t^2-\frac{1}{2}t^2\ln(t))(e^{-t})+(t \ln(t)-t)(te^{-t})\\\\\Longrightarrow\frac{1}{4}t^2e^{-t}-\frac{1}{2}t^2\ln(t)e^{-t}+ t^2\ln(t)e^{-t}-t^2e^{-t}\\\\\therefore \boxed{ y_p=\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}}$

(6) - Form the solution

$y_{gen.}=y_h+y_p\\\\\therefore\boxed{\boxed{y(t)=c_1e^{-t}+c_2te^{-t}+\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}}}$

Thus, the given DE is solved.