Question

Solve each differential equation by variation of parameters.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{lnt}$

Short answer

$y\left(t\right)=c_1{e}^{-t}+c_{2}t{e}^{-t}-\frac{3}{4}{t}^2{e}^{-t}+\frac{1}{2}{t}^2{e}^{-t}\ln t$

Step-by-step solution

Given data

The second order differential equation .$y^{\text{'}\text{'}}+2y^{\text{'}}+y=e^{-t}\ln t$

The objective is to find the solution of the differential equation by variation of parameters.

Method of variation of parameters:

Consider the differential equation,

$y^{\text{'}\text{'}}+p\left(t\right)y^{\text{'}}+q\left(t\right)y=f\left(t\right)$

Assume that $y_1\left(t\right)and{y}_2\left(t\right)$are a fundamental set of solutions for

$y^{\text{'}\text{'}}+p\left(t\right)y^{\text{'}}+q\left(t\right)y=0$

Then a particular solution to the nonhomogeneous differential equation is,

$y_p=u_1\left(x\right)y_1\left(x\right)+u_2\left(x\right)y_2\left(x\right)$

Here$u_1^{\text{'}}=\frac{W_1}{W},u_2^{\text{'}}=\frac{W_2}{W}$

$W=\left|\begin{array}{cc}{y}_1& y_2\\ y{\text{'}}_1& y{\text{'}}_2\end{array}\right|,W_1=\left|\left|\begin{array}{cc}0& y_2\\ f\left(t\right)& y{\text{'}}_2\end{array}\right|\right|and{W}_2=\left|\begin{array}{cc}{y}_1& 0\\ y{\text{'}}_1& f\left(t\right)\end{array}\right|$

Finding complementary Solution

Compare the equation (i) with$y^{\text{'}\text{'}}+2y^{\text{'}}+y=f\left(t\right)$to get $f\left(t\right)=e^{-t}\ln t$.

Auxiliary polynomial of the homogeneous equation$y^{\text{'}\text{'}}+2y^{\text{'}}+y=0$ associated with (i) is,

$\begin{array}{rcl}P\left(m\right)& =& m^2+2m+1\\ P\left(m\right)& =& 0\Rightarrow m^2+2m+1=0\\ {\left(m+1\right)}^2& =& 0\\ m& =& -1,-1\end{array}$

The roots of this polynomial are$m=-1,-1$ .

Therefore, the complementary solution is,$y_c\left(t\right)=c_1{e}^{-t}+c_{2}t{e}^{-t}$ .

Finding the Wronskian from left side

Consider$y_1\left(t\right)=e^{-t},y_2\left(t\right)=t{e}^{-t}$

Compute the Wronskian of $y_1$ and$y_2$.

$\begin{array}{rcl}W\left(y_1,y_2\right)& =& \left|\begin{array}{cc}{y}_1& y_2\\ y{\text{'}}_1& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^{-t}& t{e}^{-t}\\ -e^{-t}& (1-t)e^{-t}\end{array}\right|\\ & =& \left(1-t\right)e^{-2t}+t{e}^{-2t}\\ & =& e^{-2t}\end{array}$

Finding the Wronskian from right side of given differential quest

Compute the values of$W_1$and$W_2$:

role="math" localid="1664197518213" $\begin{array}{rcl}{W}_1& =& \left|\begin{array}{cc}0& y_2\\ f\left(t\right)& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}0& t{e}^{-t}\\ e^{-t}\ln t& (1-t)e^{-t}\end{array}\right|\\ & =& \left(0\right)\left(\left(1-t\right)e^{-t}\right)-\left(t{e}^{-t}\right)\left(e^{-t}\ln t\right)\\ & =& -t{e}^{-2t}\ln t\\ & & \end{array}$

And

$\begin{array}{rcl}{W}_2& =& \left|\begin{array}{cc}{y}_1& 0\\ y{\text{'}}_1& f\left(t\right)\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^{-t}& 0\\ -e^{-t}& e^{-t}\ln t\end{array}\right|\\ & =& \left(e^{-t}\right)\left(e^{-t}\ln t\right)-\left(0\right)\left(-e^{-t}\right)\\ & =& e^{-2t}\ln t\end{array}$

Find u'1 and u'2

Compute the derivatives as shown below:

$\begin{array}{rcl}{u}_1^{\text{'}}& =& \frac{W_1}{W}\\ & =& \frac{-t{e}^{-2t}\ln t}{e^{-2t}}\\ & =& -t\ln t\end{array}$

And

$\begin{array}{rcl}{u}_2^{\text{'}}& =& \frac{W_2}{W}\\ & =& \frac{e^{-2t}\ln t}{e^{-2t}}\\ & =& \ln t\end{array}$

Integration of u1 and u2

Calculate as shown:

$\begin{array}{rcl}{u}_1& =& \int u_1^{\text{'}}dt\\ & =& \int \left(-t\ln t\right)dt\\ & =& -\int t\ln tdt\\ & =& -\left[\left(\ln t\right)\int tdt-\int \left(\frac{d}{dt}\left(\ln t\right)\int tdt\right)dt\right]\\ & =& -\left[\frac{1}{2}{t}^2\ln t-\int \left(\frac{1}{t}·\frac{1}{2}{t}^2\right)dt\right]\\ & & \text{=}-\left[\frac{1}{2}{t}^2\ln t-\frac{1}{2}\int tdt\right]\\ & =& -\left[\frac{1}{2}{t}^2\ln t-\frac{1}{2}\left(\frac{1}{2}{t}^2\right)\right]\\ & =& -\left[\frac{1}{2}{t}^2\ln t-\frac{1}{4}{t}^2\right]\\ & =& \frac{1}{4}{t}^2\left(1-2\ln t\right)\\ & & \end{array}$

Calculate $u_2$ as shown:

$\begin{array}{rcl}{u}_2& =& \int u_2^{\text{'}}dt\\ & =& \int \left(\ln t\right)dt\\ & =& t\left(\ln t-1\right)\\ & & \end{array}$

Finding the required solution and we get

Particular solution of the differential equation (i) is as follows:

$\begin{array}{rcl}{y}_p& =& u_1{y}_1+u_2{y}_2\\ & =& \left(\frac{1}{4}{t}^2(1-2\ln t)\right)\left(e^{-t}\right)+\left(t\left(\ln t-1\right)\right)\left(t{e}^{-t}\right)\\ & =& \frac{1}{4}{t}^2{e}^{-t}\left(1-2\ln t\right)+t^2{e}^{-t}\left(\ln t-1\right)\\ & =& -\frac{3}{4}{t}^2{e}^{-t}+\frac{1}{2}{t}^2{e}^{-t}\ln t\end{array}$

Finally, general solution of the given differential equation is,

$\begin{array}{rcl}y& =& y_c+y_p\\ & =& c_1{e}^{-t}+c_{2}t{e}^{-t}-\frac{3}{4}{t}^2{e}^{-t}+\frac{1}{2}{t}^2{e}^{-t}\ln t\end{array}$