# Finding the solutions of a First order ODE

Let $y = y_{1}(t)$ be a solution of $$y' + p(t)y = 0$$

and let $y = y_{2}(t)$

be a solution of $$y' + p(t)y = g(t)$$

(a)Show that $y = y_{1}(t) + y_{2}(t)$ is also a solution of the second equation.

(b)Show that the solution of the general linear equation (the second equation) can be written in the form $y = cy_{1}(t) + y_2(t)$, where $c$ is an arbitrary constant. Identify the functions $y_{1}$ and $y_{2}$.

(c)Show that $y_1$ is a solution of the differential equation $y' + p(t)y = 0$ corresponding to $g(t) = 0$

(d) Show that $y_2$ is a solution of the full linear equation .

What i tried

(a) For the first equation i let $u(t)$ be the integrating factor. Hence $u(t)=e^{ \int^{}p(t)dt}$. I then multiply the LHS and the RHS of the equation by the integrating factor and i got $ye^{ \int^{}p(t)dt}=0$. Hence from here i deduce that $y_{1}(t)=0$.Hence, $$y = y_{1}(t) + y_{2}(t)=y_{2}(t)$$ and it is given that $y_{2}(t)$ is a solution to the second equation.

(b)I tried using the integrating factor method again to solve the equation and i got $$ye^{ \int^{}p(t)dt}=\int^{}g(t)e^{\int^{}p(t)dt}dt$$. However im stuck from here onwards as im unsure of how to change it to the form $y = cy_{1}(t) + y_2(t)$ Could anyone please explain this as well as the remaining portion of the question. Thanks

• Everything you need probably follows from this answer. Feb 10 '15 at 18:48

you don't need to know the exact form of any of the solution. what is involved the linearity of $L$ defined by $Ly = y' + py.$ this is called the principle of super position. by linearity of $L$ is meant, these two things: $$L(y_1+y_2) = Ly_1 + Ly_2, L(ky) = kLy, \text{ where k is any constant.}$$
you have $$Ly_1 = 0, Ly_2 = g.$$ by linearity of $L,$ we can conclude
$$L(y_1 + y_2) = Ly_1 + Ly_2 = 0+g=g.$$ that is, $y_1 + y_2$ is a solution of $Ly = g.$ in the same way you can show that for any $c, L(cy_1 + y_2) = g$
in other words, $cy_1 + y_2$ is a solution of $Ly = g$ for any $c.$