Chapter 4.4, Problem 1P

To determine

The general solution of the differential equation.

Answer to Problem 1P

The general solution of the initial value problem is $\underset{\_}{y=c_1+c_2\cos t+c_3\sin t+1+\ln \left|\sec t\right|-\sin t\ln \left|\sec t+\tan t\right|}$.

Explanation of Solution

Formula used:

The variation of parameters for higher order differential equations is

$v_n={\displaystyle \int \frac{W_n\left(t\right)\times g\left(t\right)}{W\left(t\right)}}dt$.

Calculation:

Consider the differential equation $y^{‴}+y^{\prime }=\tan t$, $-\frac{\pi }{2}<t<\frac{\pi }{2}$.

The characteristic polynomial for the equation is $r^3+r$.

Obtain the root of the equation as follows.

$\begin{array}{c}{r}^3+r=0\\ r\left(r^2+1\right)=0\\ r=0,\pm i\end{array}$

The roots are $r=0,\pm i$.

Therefore, the complimentary solution of the equation is $y_c=c_1{e}^{0t}+c_2{e}^{it}$.

A fundamental set of solutions $y_1,y_2\text{ and }{y}_3$

Then the general solution of the homogeneous equation is $y_c=c_1+c_2\cos t+c_3\sin t$.

Solving the fundamental set $W\left(y_1,y_2,y_3\right)\left(t\right)$ as follows.

$\begin{array}{c}W\left(y_1,y_2,y_3\right)\left(t\right)=\left|\begin{array}{ccc}{y}_1& y_2& y_3\\ {y^{\prime }}_1& {y^{\prime }}_2& {y^{\prime }}_3\\ {y^{″}}_1& {y^{″}}_2& {y^{″}}_3\end{array}\right|\\ =\left|\begin{array}{ccc}1& \cos t& \sin t\\ 0& -\sin t& \cos t\\ 0& -\cos t& -\sin t\end{array}\right|\end{array}$

Compute the latter determinant by minors as follows.

$\begin{array}{c}{W}_1\left(y_1,y_2,y_3\right)\left(t\right)=\left|\begin{array}{ccc}0& y_2& y_3\\ 0& {y^{\prime }}_2& {y^{\prime }}_3\\ 1& {y^{″}}_2& {y^{″}}_3\end{array}\right|\\ =\left|\begin{array}{ccc}0& \cos t& \sin t\\ 0& -\sin t& \cos t\\ 1& -\cos t& -\sin t\end{array}\right|\\ ={\cos }^{2}t+{\sin }^{2}t\\ =1\end{array}$

Similarly,

$\begin{array}{c}{W}_2\left(y_1,y_2,y_3\right)\left(t\right)=\left|\begin{array}{ccc}{y}_1& 0& y_3\\ {y^{\prime }}_1& 0& {y^{\prime }}_3\\ {y^{″}}_1& 1& {y^{″}}_3\end{array}\right|\\ =\left|\begin{array}{ccc}1& 0& \sin t\\ 0& 0& \cos t\\ 0& 1& -\sin t\end{array}\right|\\ =-\cos t\end{array}$

$\begin{array}{c}{W}_3\left(y_1,y_2,y_3\right)\left(t\right)=\left|\begin{array}{ccc}{y}_1& y_2& 0\\ {y^{\prime }}_1& {y^{\prime }}_2& 0\\ {y^{″}}_1& {y^{″}}_2& 1\end{array}\right|\\ =\left|\begin{array}{ccc}1& \cos t& 0\\ 0& -\sin t& 0\\ 0& -\cos t& 1\end{array}\right|\\ =-\sin t\end{array}$

By using the variation of parameters formula for obtain the particular solution.

$\begin{array}{c}{v}_1={\displaystyle \int \frac{W_1\left(t\right)\times g\left(t\right)}{1}}dt\\ ={\displaystyle \int \frac{1\times \tan t}{1}}dt\\ =\ln \left|\sec t\right|\end{array}$

Similarly,

$\begin{array}{c}{v}_2={\displaystyle \int \frac{W_2\left(t\right)\times g\left(t\right)}{1}}dt\\ ={\displaystyle \int \frac{-\cos t\times \tan t}{1}}dt\\ ={\displaystyle \int -\sin t}dt\\ =\cos t\end{array}$

$\begin{array}{c}{v}_3={\displaystyle \int \frac{W_3\left(t\right)\times g\left(t\right)}{1}}dt\\ ={\displaystyle \int \frac{-\sin t\times \tan t}{1}}dt\\ =\sin t-\ln \left|\sec t+\tan t\right|\end{array}$

Therefore, the particular solution is

$\begin{array}{c}{y}_p=v_1\left(t\right)y_1\left(t\right)+v_2\left(t\right)y_2\left(t\right)+v_3\left(t\right)y_3\left(t\right)\\ =\ln \left|\sec t\right|+{\cos }^{2}t+{\sin }^{2}t-\sin t\ln \left|\sec t+\tan t\right|\\ =1+\ln \left|\sec t\right|-\sin t\ln \left|\sec t+\tan t\right|\end{array}$

Hence, the general solution of the initial value problem is $\underset{\_}{y=c_1+c_2\cos t+c_3\sin t+1+\ln \left|\sec t\right|-\sin t\ln \left|\sec t+\tan t\right|}$.