Im trying to solve the question;
Let $y_1(t)$, $y_2(t)$ and $y_3(t)$ be solutions to the linear non homogenous equation $y’’ + p(t)y’ + q(t)y = g(t)$, where $p(t), q(t)$ and $g(t)$ are continuous on $ℝ$, such that $y_1(0) = y_1’(0) = 0$, $y_2(0) = 0, y_2’(0) = 1$, $y_3(0) = 1, y_3’(0) = 0$. What is the general solution of this equation? What is the solution of the equation satisfying the initial conditions $y(0) = 2$ and $y’(0) = 3$?
I believe two of $y_1(t)$, $y_2(t)$ and $y_3(t)$ should be the solution to the homogeneous equation $y’’ + p(t)y’ + q(t)y =0$, and one of them should be the particular solution. But I'm not sure how to continue from there.
Any help would be appreciated. Thanks in advance.
Best Answer
Your reasoning is correct. For the general solution of the equation, one of $y_{1}(t),y_{2}(t)$ or $y_{3}(t)$ is the particular solution and the other two are the solutions to the homogenous equation. However, you need more information to determine them, which are the given initial conditions. The particular solution to the non-homogenous equation does not contain any arbitrary constants (see this), but the homogenous solutions do.
Looking at the given initial conditions, it is clear that $y_{1}(0)$ and $y_1'(0)$ do not contribute to the initial conditions $y(0)$ and $y'(0)$, while $y_{2}(t)$ and $y_{3}(t)$ do. Thus, $y_{2}(t)$ and $y_{3}(t)$ are solutions to the homogenous equation and $y_{1}(t)$ is a particular solution. The general solution is then
$$y(x)=Ay_{2}(t)+By_{3}(t)+y_{1}(t)$$
Now substituting the initial conditions, you can find the constants $A$ and $B$.