Solving system of differential equations using elimination method

[berlinvic]

Homework Statement

$$\begin{cases}
y'_1 = y_2, \\
y'_2 = -y_1 + \frac{1}{\cos x}
\end{cases}$$

Relevant Equations

Elimination method for differential equations, associated homogeneous equation

I am trying to solve this system of differential equations using elimination method, but I am stuck.

$$\begin{cases}
y'_1 = y_2, \\
y'_2 = -y_1 + \frac{1}{\cos x}
\end{cases}$$

Here's what I tried:

I've been suggested to differentiate the $y_1'= y_2$ again to get $y_1''= y_2'= -y_1+ \frac{1}{\cos(x)}$. The "associated homogeneous equation". However, I don't really see a way how to go forward. Anyways, how do you solve this sytem?

 

[Orodruin]

That’s just a second order linear homogeneous differential equation. It can be solved by any appropriate method.

 

[Mark44]

I've been suggested to differentiate the $y_1'= y_2$ again to get $y_1''= y_2'= -y_1+ \frac{1}{\cos(x)}$. The "associated homogeneous equation".

That’s just a second order linear homogeneous differential equation.

Wouldn't the linear homogeneous diff. equation be $y_1'' + y_1 = 0$? The $\frac 1 {\cos(x)}$ term makes it nonhomogeneous.

 

[Orodruin]

Wouldn't the linear homogeneous diff. equation be $y_1'' + y_1 = 0$? The $\frac 1 {\cos(x)}$ term makes it nonhomogeneous.

Yes sorry, I meant to say inhomogeneous …

 

[berlinvic]

Yes sorry, I meant to say inhomogeneous …

But I don't see how I can solve $y_1^{\prime\prime}=y_2^{\prime}=-y_1+\frac1{\cos(x)}$ using elimination method. Any help? I get the same answer as using https://mathdf.com/dif/, which, according to my professor, is incorrect.

 

[Orodruin]

You already eliminated $y_2$ ….

 

[berlinvic]

Yes, but I still can't see how to move forward, I am just stuck at this line.

 

[Mark44]

@berlinvic, did you get a solution for the homogeneous problem $y_1'' + y_1 = 0$? That's an easy one. For the nonhomogeneous problem $y_1'' + y_1 = \sec(x)$, you might have to use variation of parameters.

 

[berlinvic]

Yes, I should have mentioned, I already know I should use variation of parameters. However I am not entirely sure which ones and how to apply them. I am fairly confused about it, do you have any tips?

 

[berlinvic]

Here is everything I've got so far:

I reduced the system to the second order ODE
$$
y_1''+y_1=\frac{1}{\cos x}. \tag{1}
$$

Next steps:
1. Find two linearly independent solutions to the homogeneous equation $y_1''+y_1=0$ --- for instance, $u(x)=\cos x$ and $v(x)=\sin x$;
2. Compute the Wronskian of those solutions:
$$
W[u,v]=uv'-u'v; \tag{2}
$$
3. Find a particular solution to the equation $(1)$ using the [formula](https://en.wikipedia.org/wiki/Variation_of_parameters#General_second-order_equation)
$$
y_{1p}(x)=A(x)u(x)+B(x)v(x), \tag{3}
$$
where
$$
A(x)=-\int\frac{v(x)}{W(x)\cos x}\,dx,\qquad
B(x)=\int\frac{u(x)}{W(x)\cos x}\,dx; \tag{4}
$$
4. The general solution to $(1)$ is given by
$$
y_1(x)=c_1u(x)+c_2v(x)+y_{1p}(x); \tag{5}
$$
5. To find $y_2(x)$, use the first equation of the system:
$$
y_2(x)=y_1'(x)=c_1u'(x)+c_2v'(x)+y_{1p}'(x). \tag{6}$$

 

[Mark44]

Here is everything I've got so far:

I reduced the system to the second order ODE
$$
y_1''+y_1=\frac{1}{\cos x}. \tag{1}
$$

Next steps:
1. Find two linearly independent solutions to the homogeneous equation $y_1''+y_1=0$ --- for instance, $u(x)=\cos x$ and $v(x)=\sin x$;
2. Compute the Wronskian of those solutions:
$$
W[u,v]=uv'-u'v; \tag{2}
$$

What did you get for the Wronksian? It's pretty simple in this problem.

3. Find a particular solution to the equation $(1)$ using the [formula](https://en.wikipedia.org/wiki/Variation_of_parameters#General_second-order_equation)
$$
y_{1p}(x)=A(x)u(x)+B(x)v(x), \tag{3}
$$
where
$$
A(x)=-\int\frac{v(x)}{W(x)\cos x}\,dx,\qquad
B(x)=\int\frac{u(x)}{W(x)\cos x}\,dx; \tag{4}
$$

I think you're confusing yourself with the above notation. The same wiki page you quoted gives these formulas for A(x) and B(x):
$A(x) = -\int \frac 1 W u_1(x)f(x)dx$
$B(x) = \int \frac 1 W u_2(x)f(x)dx$
Here W is the Wronskian and $u_1(x)$ and $u_2(x)$ are the solutions to the homogeneous DE you found earlier.

4. The general solution to $(1)$ is given by
$$
y_1(x)=c_1u(x)+c_2v(x)+y_{1p}(x); \tag{5}
$$
5. To find $y_2(x)$, use the first equation of the system:
$$
y_2(x)=y_1'(x)=c_1u'(x)+c_2v'(x)+y_{1p}'(x). \tag{6}$$