Author Topic: Q4 TUT 0801  (Read 1024 times)

Victor Ivrii

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Q4 TUT 0801
« on: October 26, 2018, 05:44:36 PM »
Find the general solution of the given differential equation.
$$
4y'' + y = 2 \sec(t/2),\qquad -\pi  < t < \pi.
$$

Monika Dydynski

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Re: Q4 TUT 0801
« Reply #1 on: October 26, 2018, 07:34:19 PM »
Find the general solution of the given differential equation.

$$4y'' + y = 2 \sec\left(\frac{t}{2}\right)\tag{1},\qquad -\pi  < t < \pi.$$

Dividing $(1)$ by $4$, we have

$$y''+\frac{1}{4}y=\frac{1}{2}\sec{\frac{t}{2}},$$


where $p(t)=0$, $q(t)=\frac{1}{4}$, and $g(t)=\frac{1}{2}\sec{\frac{t}{2}}$ are continuous on $(-\pi, \pi)$.

The corresponding homogeneous equation is
$$4y''+y=0,$$

with characteristic equation
$$4r^2+1=0 \Rightarrow r_{1,2}=\pm\frac{i}{2}.$$

The homogeneous solution is $y_h(t)=c_1\cos\frac{t}{2}+c_2\sin\frac{t}{2}$


Computing the Wronskian,


$$W\left(\cos{\frac{t}{2}}, \sin{\frac{t}{2}}\right)(t)=\begin{array}{|c c|}\cos{\frac{t}{2}}& \sin{\frac{t}{2}} \\ -\frac{1}{2}\sin{\frac{t}{2}} &\frac{1}{2}\cos{\frac{t}{2}}\end{array}=\frac{1}{2}\ne0,$$

we verify that $y_1(t)=\cos{\frac{t}{2}}$ and $y_2(t)=\sin{\frac{t}{2}}$ form a fundamental set of solutions.


Calculating the parameters $u_1$ and $u_2$, we have

$$u_1=-\int{\frac{y_2(t)g(t)}{W(y_1,y_2)(t)}}dt=-\int{\tan{\frac{t}{2}}}dt=2\ln\left(\cos{\frac{t}{2}}\right)+c_1$$

$$u_2=\int{\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}}dt=\int{1}dt=t+c_2.$$

It follows that the general solution is

$$y(t)=\cos{\frac{t}{2}}\left(2\ln{\left(\cos{\frac{t}{2}}\right)}+c_1\right)+\sin{\frac{t}{2}}(t+c_2)=2\cos{\frac{t}{2}}\ln{\left(\cos{\frac{t}{2}}\right)}+t\sin{\frac{t}{2}}+c_1\cos{\frac{t}{2}}+c_2\sin{\frac{t}{2}}.$$





« Last Edit: October 26, 2018, 07:44:36 PM by Monika Dydynski »