Solving system of differential equations using elimination

Question

I am trying to solve this system of differential equations using elimination method, but I am stuck. $$\left.\left\{\begin{aligned}&y_1^{\prime}=y_2,\\&y_2^{\prime}=-y_1+\frac1{\cos x}.\end{aligned}\right.\right.$$ My attempt:

I tried differentiating the $y_1'= y_2$ again to get $y_1''= y_2'= -y_1+ \frac{1}{\cos(x)}$. But I am not sure how to proceed further. I've been suggested to use Lagrange method later, but I am not sure. I would appreciate if someone could help me get to the final solution.

Answer 1

You 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 $$ y_{1p}(x)=A(x)u(x)+B(x)v(x), \tag{3} $$ where \begin{align} A(x)&=-\int\frac{v(x)}{W(x)\cos x}\,dx,\tag{4a} \\ B(x)&=\int\frac{u(x)}{W(x)\cos x}\,dx; \tag{4b} \end{align}
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: $$ y_2(x)=y_1'(x)=c_1u'(x)+c_2v'(x)+y_{1p}'(x). \tag{6} $$

Answer 2

What makes you think you can proceed further?

In this special case, due to the differential equation of the cosine, you can indeed integrate one step ahead: $$ 1=uy''-u''y=(uy'-u'y)' \\ \implies x+c = uy'-u'y=\cos(x)y'(x)+\sin(x)y(x) $$ In principle this is a linear DE of first order, you should know the solution algorithm or formula. The resulting terms will not be nice.