How to use the Wronskian to find the in homogeneous solution of an ODE?

Suppose I have an ODE in the form:

$y'' + p(x)y' + q(x)y = f(x)$

Suppose now that I have obtained the solution to the equation

$y'' + p(x)y' + q(x)y = 0$

which is in the form

$y(x) = c_1 y_1(x) + c_2 y_2(x)$

My textbook suggests that it is possible to use the Wronskian of the solution in order to obtain a solution to the inhomogeneous equation, hence getting the general solution to the equation.

For example, the homogeneous solution to:

$y'' + 6y' + 2y = 1$

is $c_1exp(-3+\sqrt 7)x + c_2exp(-3-\sqrt 7)x$

How is then one able to use the Wronskian to obtain the general solution (which I am able to derive ($y_p = 0.5$) by other methods, but not this one).

• use the formula that gives the particular solution in function of the wronskian – Isham Mar 26 '18 at 18:44
• What would that formula be? – daljit97 Mar 26 '18 at 18:58
• Saheb has posted it – Isham Mar 26 '18 at 19:05

You can use method of variation of parameters in such cases. The general solution is $$y(x)=p(x)y_1(x)+q(x)y_2(x),$$ where $p(x)=-\int \dfrac{y_2(x)f(x)}{w(y_1,y_2)} dx+c_1$ and $q(x)=\int \dfrac{y_1(x)f(x)}{w(y_1,y_2)} dx+c_2$ and $w(y_1,y_2)$ be the Wronskian of $y_1(x)$ and $y_2(x)$.

• I am familiar with the variation of parameters, I usually solve the simultaneous equations: $\begin{cases} g_1'u_1 + g_2'u_2 = 0\\ g_1'u_1' + g_2'u_2' = f(x) \end{cases}$ where $u_1$ and $u_2$ are the solutions of the homogeneous equation. $\\$ Is that the equivalent of what you suggested? I am asking because my textbook says "Use the Wronskian method to show that the ordinary differential equation" – daljit97 Mar 26 '18 at 19:13
• Yes. This is equivalent. Solve for $g_1'$ and $g_2'$ and see what happens. – SAHEB PAL Mar 26 '18 at 19:17
• Ok what does $f(x)$ refers to? – daljit97 Mar 26 '18 at 23:51
• $f(x)$ refers to the nono-homogeneous part of your ODE . – SAHEB PAL Mar 27 '18 at 6:39