Question

Solve each differential equation by variation of parameters.

${\mathbf{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

Short answer

$y=k_1{e}^t+k_{2}t{e}^t+\frac{1}{2}{t}^2{e}^t{\tan }^{-1}\left(t\right)-\frac{1}{2}{e}^t{\tan }^{-1}\left(t\right)-\frac{1}{2}t{e}^t\ln \left(1+t^2\right)$

Step-by-step solution

Note the given data

Given the differential equation$y^{\text{'}\text{'}}-2y^{\text{'}}+y=e^t{\tan }^{-1}\left(t\right)$.$y^{\text{'}\text{'}}-2y^{\text{'}}+y=e^t{\tan }^{-1}\left(t\right)$

We need to find the general solution of the given differential equation

Finding the solution of the homogeneous solution

Finding the solution of the homogeneous equation

$y^{\text{'}\text{'}}-2y^{\text{'}}+y=0$…………………………………………………………..(1)

Let $y=e^{mx}$be a solution of the given homogeneous solution

Differentiate the above equation with respect to x , we get

$y^{\text{'}}=m{e}^{mx}$………………………………………………………………(2)

Again, differentiate the above equation with respect to x , we get

$y^{\text{'}}=m{e}^{mx}$…………………………………………………………….(3)

Substitute the value$y=e^{mx}$and equation (3) in differential equation (1), we get

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

Since $e^{mx}\ne 0$, then we get the roots

$m=\pm i$

Therefore, the roots are complex and conjugate.

Then the solution of a homogeneous solution is

$y_h=c_1{y}_1+c_2{y}_2$

$=c_1{e}^t+c_{2}t{e}^t$………………………………………………………………..(4)

Finding the Wronskian from left side

We need to find the particular solution

Since we have$y_1=e^{-z}$and $y_2=e^{-2x}$, then we can obtain the wronskian as

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

Finding the Wronskian from right side of given differential question

Now we find the Wronskian$W_1\text{and}{W}_2$ of right side of the differential equation

$f\left(x\right)=\sec x$

$\begin{array}{rcl}{W}_1& =& \left|\begin{array}{cc}0& y_2\\ f\left(t\right)& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}0& t{e}^t\\ e^t{\tan }^{-1}\left(t\right)& (t+1)e^t\end{array}\right|\\ & =& 0-\left(t{e}^t\right)\times \left(e^t{\tan }^{-1}\left(t\right)\right)\\ & =& -t{e}^{2t}{\tan }^{-1}\left(t\right)\end{array}$

And

$\begin{array}{rcl}{W}_2& =& \left|\begin{array}{cc}{y}_1& 0\\ y{\text{'}}_1& f\left(t\right)\end{array}\right|\\ & =& \left|\begin{array}{cc}0& t{e}^t\\ e^t{\tan }^{-1}\left(t\right)& (t+1)e^t\end{array}\right|\\ & =& 0-\left(t{e}^t\right)\times \left(e^t{\tan }^{-1}\left(t\right)\right)\\ & =& -t{e}^{2t}{\tan }^{-1}\left(t\right)\end{array}$

Finding the value of u1 and u2

Now, we find the value of

$\begin{array}{rcl}{u}_1^{\text{'}}& =& \frac{W_1}{W\left(y_1,y_2\right)}\\ & =& \frac{-t{e}^{2t}{\tan }^{-1}\left(t\right)}{e^{2t}}\\ & =& -e^{-2t}\times t{e}^{2t}{\tan }^{-1}\left(t\right)\\ & =& -t{\tan }^{-1}\left(t\right)\end{array}$

And

$\begin{array}{rcl}{u}_2^{\text{'}}& =& \frac{W_2}{W\left(y_1,y_2\right)}\\ & =& \frac{e^{2t}{\tan }^{-1}\left(t\right)}{e^{2t}}\\ & =& e^{-2t}\times e^{2t}{\tan }^{-1}\left(t\right)\\ & =& {\tan }^{-1}\left(t\right)\end{array}$

By integration, we get

$\begin{array}{rcl}{u}_1& =& \int u_1^{\text{'}}dt\\ & =& -\int t{\tan }^{-1}\left(t\right)dt\end{array}$

And

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

Finding the required solution and we get

Substitute all the values of ,$u_1\text{and}{u}_2\text{and the functions}{y}_1\text{and}{y}_2$ we get the particular solution

$\begin{array}{rcl}y& =& y_h+y_p\\ & =& c_1{e}^t+c_{2}t{e}^t+\frac{1}{2}t{e}^t-\frac{1}{2}{e}^t{\tan }^{-1}\left(t\right)+\frac{1}{2}{t}^2{e}^t{\tan }^{-1}\left(t\right)-\frac{1}{2}t{e}^t\ln \left(1+t^2\right)\\ & =& c_1{e}^t+\left(c_2+\frac{1}{2}\right)t{e}^t+\frac{1}{2}{t}^2{e}^t{\tan }^{-1}\left(t\right)-\frac{1}{2}{e}^t{\tan }^{-1}\left(t\right)-\frac{1}{2}t{e}^t\ln \left(1+t^2\right)\\ & =& k_1{e}^t+k_{2}t{e}^t+\frac{1}{2}{t}^2{e}^t{\tan }^{-1}\left(t\right)-\frac{1}{2}{e}^t{\tan }^{-1}\left(t\right)-\frac{1}{2}t{e}^t\ln \left(1+t^2\right)\end{array}$

Now, we find the general solution

.$y=k_1{e}^t+k_{2}t{e}^t+\frac{1}{2}{t}^2{e}^t{\tan }^{-1}\left(t\right)-\frac{1}{2}{e}^t{\tan }^{-1}\left(t\right)-\frac{1}{2}t{e}^t\ln \left(1+t^2\right)$