Solving C1, C2 for linear homogenous DE

Question

I have the following question:\begin{align*} y_1&= 2\sin x+3\cos x\\ y_2 &= \sin x+2\cos x \end{align*} forms a fundamental set of solutions to the second order linear homogenous equation $y{''} + y = 0$ on $(−\infty, +\infty)$. One can also check that ̃$\tilde {y} = \sin x$ is a solution to this differential equation. One should be able to express $\sin x$ as $c_{1}*y_{1}+ c_{2}*y_{2}$ for some constants c1, c2. Find such c1, c2.

I thought to write out the Wronskian of $y_1$ and $y_2$ and then plug $x=0$ into the Wronskian but I am not sure what the other side of the system of equations would be. Could anyone help me figure this out? That is what $c_{1}*y_{1}+ c_{2}*y_{2}$ and $c_{1}'*y_{1}+ c_{2}'*y_{2}$ would be equal to at $x=0$. I could then invert the matrix and solve for the solution, $c_1$ $c_2$.

Any thoughts and comments are greatly appreciated. Thank you.

Answer 1

For the sum to be $\sin x$, you want the coefficients of the sines to add to $1$, and the coefficients of the cosines to add to $0$.

This means that $$c_1(2\sin x+3\cos x)+c_2(\sin x+2\cos x)=1 \sin x + 0 \cos x,$$ or, after expanding and regrouping, \begin{align}2c_1 \sin x +3 c_1 \cos x+c_2 \sin x+2 c_2 \cos x&=1\sin x+0\cos x\\ (2c_1+c_2)\sin x+(3 c_1+2 c_2)\cos x&=1 \sin x+0\cos x.\end{align}

Equating the coefficients of $\sin x$ and $\cos x$ on the LHS and RHS in the above equation gives you $$2 c_1+c_2=1$$ $$3 c_1+2c_2=0,$$

a system of equations which you can solve to get $c_1$ and $c_2$.

Answer 2

You have to solve $2c_1+c_2=1$ and $3c_1+2c_2=0$. This is a system of two equation which can be solved. You get $c_1=2$ and $c_2=-3$.

Answer 3

The general solution is $$\begin{align*} y&= 2c_1\sin x+3c_1\cos x\\ &+ c_2\sin x+2c_2\cos x \end{align*}$$ Solve the following system $$\begin{align*} 1&= 2c_1+c_2\\ 0&= 3c_1+2c_2 \end{align*} \implies \begin{cases} c_2=1-2c_1\\ c_1=2 \end{cases}\implies \begin{cases} c_2=-3\\ c_1=2 \end{cases}$$

Answer 4

We can write the definitions of $y_1$ and $y_2$ in matrix form: $$\begin{pmatrix}y_1\\ y_2\end{pmatrix}=\underbrace{\begin{pmatrix}2&3\\1&2\end{pmatrix}}_P\begin{pmatrix}\sin x\\\cos x\end{pmatrix}.$$ $P$ is invertible ($\det P=-1$) so we know $(y_1, y_2)$ is a basis of the space of solutions and $$\begin{pmatrix}\sin x\\\cos x\end{pmatrix}=P^{-1}\begin{pmatrix}y_1\\ y_2\end{pmatrix}.$$ Since $P is a $2\times 2$ matrix, it's simples to find the inverse matrix using the adjugate matrix: $$\operatorname{adj}(P)=\begin{pmatrix}2&-3\cr -1&2\end{pmatrix},\quad \text{so }\enspace P^{-1}=\frac1{\det P}~\operatorname{adj}(P)=\begin{pmatrix}-2&3\cr 1&-2\end{pmatrix}$$ $$\text{and }\qquad\begin{cases}\sin x=-2y_1+3y_2,\cr\cos x=\phantom{-2}y_1-2y_2.\end{cases}$$